From 7f9172f15e0e076508c8ed25a226596d4dba6b4a Mon Sep 17 00:00:00 2001 From: semapheur Date: Thu, 9 May 2024 22:53:22 +0200 Subject: [PATCH] Deploy --- content/notes/math/linear_algebra.mdx | 583 ++++++++++++++++++++- content/notes/math/manifolds.mdx | 58 +- content/notes/math/measure_theory.mdx | 6 +- content/notes/math/stochastic_analysis.mdx | 4 +- 4 files changed, 616 insertions(+), 35 deletions(-) diff --git a/content/notes/math/linear_algebra.mdx b/content/notes/math/linear_algebra.mdx index 20c8b66..0b7ec21 100644 --- a/content/notes/math/linear_algebra.mdx +++ b/content/notes/math/linear_algebra.mdx @@ -788,7 +788,7 @@ $$ which is subspace of the vector space of all functions from $I$ to $\bigcup_{i\in I} V_i$. The support of $f:I\to\bigcup_{i\in I} V_i$ is the set $$ - \mathrm{supp}(f) = \Set{i\in I | f(i) \neq 0} + \operatorname{supp}(f) = \Set{i\in I | f(i) \neq 0} $$ Thus, $f$ has finite support if $f(i) = 0$ for all but a finite number of $i\in I$. @@ -3473,6 +3473,18 @@ and if the ranks are all finite, then $$ \operatorname{rank}(M\setminus S) = \operatorname{rank}(M) - \operatorname{rank}(S) $$ + +
+Proof + +If $f: M\to F$ is a module epimorphism where $F$ is free on $B$, then it is easy to define a right inverse for $f$, since we can define an $R$-map $f_R: F\to M$ by specifying its values arbitrarily on $B$ and extending by linearity. Thus, we take $f_R (\mathbf{b})$ to be any member of $f^{-1}(\mathbf{b})$. Then by a previous theorem, $\ker(f)$ is a direct summand of $M$ and + +$$ + M \cong \ker(f)\boxplus F +$$ + +This applies to the canonical projection $\pi:M\to M\setminus S$ provided that the quotient $M\setminus S$ is free. +
## Quotient modules @@ -3697,6 +3709,575 @@ The following is a list of some of the properties of modules over commutative ri Recall that a module over a noncommutative ring may have bases of different sizes. However, all bases for a free module over a commutative ring with identity have the same size. +# Modules over a principal ideal domain + +## Annhilators and orders + +When $R$ is a principal ideal domain, all annihilators are generated by a single element. + + +Let $R$ be a principal ideal domain and let $M$ be an $R$-module. +1. If $N$ is a submodule of $M$, then any generator of $\operatorname{ann}(N)$ is called an *order* of $N$. +2. An *order* of an element $\mathbf{v}\in M$ is an order of the submodule $\langle \mathbf{v} \rangle$. + + + +Let $R$ be a principal ideal domain and let $M$ be an $R$-module. +1. If $\alpha$ is an order of $N\leq M$, then the orders of $N$ are precisely the associates of $\alpha$. We denote any order of $N$ by $o(N)$. +2. If $M = A \oplus B$, then $o(M) = \operatorname{lcm}(o(A),o(B))$, i.e. the orders of $M$ are precisely the least common multiples of the orders of $A$ and $B$. + +
+Proof + +**(2):** Suppose that + +$$ +\begin{align*} + o(M) =& \delta \\ + o(A) =& \alpha \\ + o(B) =& \beta \\ + \lambda =& \operatorname{lcm}(\alpha, \beta) +\end{align*} +$$ + +Then $\delta A = \Set{0}$ and $\beta B = \Set{0}$ imply that $\alpha | \delta$ and $\beta | \delta$ and so $\lambda | \delta$. On the other hand, $\lambda$ annihilates both $A$ and $B$ and therefore also $M = A \oplus B$. Hence, $\delta | \lambda$ and so $\lambda\sim\delta$ is an order of $M$. +
+ + +## Cyclic modules + + +Let $R$ be a principal ideal domain. +1. If $\langle\mathbf{v}\rangle$ is a cyclic $R$-module with annihilator $\langle\alpha\rangle$, then the multiplication map $f:R\to\langle\mathbf{v}\rangle$ defined by $f\rho = \rho\mathbf{v}$ is an $R$-epimorphism with $\ker(f) = \langle\alpha\rangle$. Hence, the induced map $\bar{g}: R\setminus\langle\alpha\rangle \to \langle\mathbf{v}\rangle$ defined by $\bar{g}(\rho + \langle\alpha\rangle) = \rho\mathbf{v}$ is an isomorphism. In other words, cyclic $R$-modules are isomorphic to quotient modules of the base ring $R$. +2. Any submodule of a cyclic $R$-module is cyclic. +3. If $\langle\mathbf{v}\rangle$ is a cyclic submodule of $M$ of order $\alpha$, then for $\beta\in R$, +$$ + o(\langle\beta\mathbf{v}\rangle) = \frac{\alpha}{\gcd(\beta,\alpha)} +$$ +Also +$$ +\begin{align*} + &\langle\beta\mathbf{v}\rangle = \langle\mathbf{v}\rangl \\ + \iff& (o(\mathbf{v}),\beta) = 1 \\ + \iff& o(\beta\mathbf{v}) = o(\mathbf{v}) +\end{align*} +$$ + +
+Proof + +**(2):** Let $S\leq\langle\mathbf{v}\rangle$, then $\mathcal{I} = \Set{\rho\in R | \rho\mathbf{v}\in S}$ is an ideal of $R$ and so $\mathcal{I} = \langle\sigma\rangle$ for some $\sigma\in R$. Thus, + +$$ + S = \mathcal{I}\mathbf{v} = R\sigma\mathbf{v} = \langle\sigma\mathbf{v}\rangle +$$ + +**(3):** We have $\rho(\beta\mathbf{v}) = 0$ if and only iff $\rho(\beta)\mathbf{v} = 0$, i.e. if and only if $\alpha|\rho\beta$, which is equivalent to + +$$ + \gamma := \left.\frac{\alpha}{\gcd(\alpha,\beta)}\right| r +$$ + +Thus, $\rho\in\operatorname{ann}(\beta\mathbf{v})$ if and only if $\rho\in\langle\gamma\rangle$ and so $\operatorname{ann}(\beta\mathbf{v}) = \langle\gamma\rangle$. For the second statements, if $(alpha,\beta) = 1$ then there exist $a,b \in R$ for which $a\alpha + b\beta = 1$ and so + +$$ + \mathbf{v} = (a\alpha + b\beta)\mathbf{v} = b\beta\mathbf{v} \in\langle\beta\mathbf{v}\rangle \subseteq \langle\mathbf{v}\rangle +$$ + +and so $\langle\beta\mathbf{v}\rangle = \langle\mathbf{v}\rangle$. Indeed, if $\langle\beta\mathbf{v}\rangle = \langle\mathbf{v}\rangle$, then $o(\beta\mathbf{v}) = \alpha$. Finally, if $o(\beta\mathbf{v}) = \alpha$, then + +$$ + \alpha = o(\beta\mathbf{v}) = \frac{\alpha}{\gcd(\alpha,\beta)} +$$ + +and so $(\alpha,\beta) = 1$. +
+ + +### Decomposition of cyclic modules + + +Let $M$ be an $R$-module. +1. **Composing cyclic modules:** If $u_1,\dots,u_n \in M$ have relatively prime orders, then $o\left(\sum_{i=1}^n \mathbf{u}_i \right) = \prod_{i=1}^n o(\mathbf{u}_i)$ and $\bigoplus_{i=1}^n \langle\mathbf{u}_i \rangle = \left\langle \sum_{i=1}^n \mathbf{u}_i \right\rangle$. Consequently, if $M = \sum_{i=1}^n A_i$ where the submodules $A_i$ have relatively prime orders, then the sum is direct. +2. **Decomposing cyclic modules:** If $o(\mathbf{v}) = \prod_{i=1}^n \alpha_i$ where the $\alpha_i$ are pairwise relatively prime, then $\mathbf{v}$ has the form $\mathbf{v} = \sum_{i=1}^n \mathbf{u}_i$ where $o(\mathbf{u}_i) = \alpha_i$ and so +$$ + \langle\mathbf{v}\rangle = \left\langle \sum_{i=1}^n \mathbf{u}_i \right\rangle = \bigoplus_{i=1}^n \langle\mathbf{u}_i \rangle +$$ + +
+Proof + +**(1):** Let $\alpha_k = o(\mathbf{u}_k)$, $\mu := \prod_{i=1}^n \alpha_i$ and $\mathbf{v} := \sum_{i=1}^n \mathbf{u}_i$. Since $\mu$ annihilates $\mathbf{v}$, the order of $\mathbf{v}$ divides $\mu$. If $o(\mathbf{v})$ is a proper divisor of $\mu$, then for some index $k$, there is a prime $p|\alpha_k$ for which $\frac{\mu}{p}$ annihilates $\mathbf{v}$. However, $\frac{\mu}{p}$ annihilates each $\mathbf{u}_i$ for $i \neq k$. Thus + +$$ + 0 = \frac{\mu}{p}\mathbf{v} = \frac{\mu}{p}\mathbf{u}_k = \frac{\alpha_k}{p}\left(\frac{\mu}{p}\right)\mathbf{u}_k +$$ + +Since $o(\mathbf{u}_k)$ and $\frac{\mu}{\alpha_k}$ are relatively prime, the order of $\frac{\mu}{\alpha_k}\mathbf{u}_k$ is equal to $o(\mathbf{u}_i) = \alpha_k$, which contradicts the equation above. Hence, $o(\mathbf{v}) = \mu$. + +It is clear that $\left\langle \sum_{i=1}^n \mathbf{u}_i \right\rangle \subseteq bigoplus_{i=1}^n \langle\mathbf{u}_i \rangle$. For the reverse inclusion, since $\alpha_1$ nad $\frac{\mu}{\alpha_1}$ are relatively prime, there exist $r,s \in R$ for which + +$$ + r\alpha_1 + s\frac{\mu}{\alpha_1} = 1 +$$ + +Hence + +$$ +\begin{align*} + \mathbf{u}_1 =& \left( r\alpha_1 + s\frac{\mu}{\alpha_1} \right)\mathbf{u}_1 = s\frac{\mu}{\alpha_1} \\ + =& s \frac{\mu}{\alpha_1} \sum_{i=1}^n \mathbf{u}_n \in \left\langle\sum_{i=1}^n \mathbf{u}_i \right\rangle +\end{align*} +$$ + +Similarly, $\mathbf{u}_k \in\left\langle\sum_{i=1}^n \mathbf{u}_i \right\rangle$ for all $k$ and so we get the reverse inclusion. Finally, to see that the sum above is direct, note that if $\sum_{i=1}^n \mathbf{v}_i = 0$, where $\mathbf{v}_i \in A_i$, then each $\mathbf{v}_i$ must be $0$, for otherwise the order of the sum on the left would be different from $1$. + +**(2):** The scalars $\beta_k = \frac{\mu}{\alpha_k}$ are relatively prime and so there exist $a_i \in R$ for which $\sum_{i=1}^n a_i \beta_i = 1$. Hence, + +$$ + \mathbf{v} = \left(\sum_{i=1}^n a_i \beta_i \right)\mathbf{v} = \sum_{i=1}^n a_i \beta_i \mathbf{v} +$$ + +Since $o(\beta_k \mathbf{v}) = \frac{\mu}{\gcd(\mu,\beta_k)} = \alpha_k$ and since $a_k$ and $\alpha_k$ are relatively prime, we have $o(a_k \beta_k \mathbf{v}) = \alpha_k$. The second statement follows from **(1)**. +
+ + +## Free modules of a prinicipal ideal domain + + +Let $M$ be a free module over a principal ideal domain $R$. Then any submodule $S$ of $M$ is also free and $\operatorname{rank}(S) \leq\operatorname{rank}(M)$. + +
+Proof + +We will give the proof first for modules of finite rank and then generalize to modules of arbitrary rank. Since $M\cong R^n$ where $n = \operatorname{rank}(M)$ is finite, we may assume that $M = R^n$. For each $1\leq k \leq n$, let + +$$ + \mathcal{I} = \Set{\rho\in R | (\alpha_1,\dots,\alpha_{k-1},\rho,0,\dots,0)\in S,\; \alpha_1,\dots,\alpha_{k-1}\in R} +$$ + +Then it is easy to see that $\mathcal{I}_k$ is an ideal of $R$ and so $\mathcal{I}_k = \langle \rho_k \rangle$ for some $\rho_k \in R$. Let + +$$ + \mathbb{u}_k = (\alpha_1,\dots,\alpha_{k-1},\rho_k,0,\dots,0)\in S +$$ + +We claim that $B = \Set{\mathbf{u}_k | \rho_k \neq 0}_{k=1}^n$ is a basis for $S$. As to linear independence, suppose that $B = \Set{\mathbf{u}_{i_j}}_{j=1}^m$ and that $\sum_{\ell=1}^s \alpha_{j_\ell}\mathbf{u}_{j_\ell}$. Comparing the $j_s$th coordinates gives $\alpha_{j_s}\rho_{j_s} = 0$ and since $\rho_{j_s} \neq 0$, if follows that $\alpha_{j_s} = 0$. In a similar way, all coefficients are $0$ and so $B$ is linearly independent. + +To see that $B$ spans $S$, we partition the elements $\mathbf{x}\in S$ according to the largest coordinate index $i(\mathbf{x})$ with nonzero entry and induct on $i(\mathbf{x})$. If $i(\mathbf{x}) = 0$, then $\mathbf{x} = 0$, which is in the span of $B$. Suppose that all $\mathbf{x}\in S$ with $i(\mathbf{x}) < k$ are in the span of $B$ and let $i(\mathbf{x}) = k$, i.e. + +$$ + \mathbf{x} = (\alpha_1,\dots,\alpha_k,0,\dots,0),\; \alpha_k \neq 0 +$$ + +Then $\alpha_k \in\mathcal{I}_k$ and so $\rho_k \neq 0$ and $\alpha_k = c\rho_k$ for some $c\in R$. Thus, $i(\mathbf{x} - c\mathbf{u}_k) < k$ and so $\mathbf{y} = \mathbf{x} - c\mathbf{u}_k \in \langle B \rangle$ and therefore $\mathbf{x}\in\langle B \rangle$. Hence, $B$ is a basis for $S$. + +The proof can be generalized to modules of arbitrary rank. In this case, we may assume that $M = (R^\kappa)_0$ is the $R$-module of functions with finite support from $\kappa$ to $R$, where $\kappa$ is a cardinal number. We use that fact that $\kappa$ is a well-ordered set, i.e. $\kappa$ is a totally ordered set in which any nonempty subset has a smallest element. If $\alpha \in \kappa$, the closed interval $[0,\alpha]$ is + +$$ + [0,\alpha] = \Set{x \in\kappa | 0 \leq x \leq \alpha} +$$ + +Let $S\leq M$. For each $0 < \alpha \leq\kappa$, let + +$$ + M_\alpha = \Set{f\in S | \operatorname{supp}(f) \subseteq [0,\alpha]} +$$ + +Then the set $\mathcal{I}_\alpha = \Set{f(\alpha) | f\in M_\alpha}$ is an ideal of $R$ and so $\mathcal{I}_\alpha = \langle \mathbf{f}_\alpha (\alpha) \rangle$ for some $\mathbf{f}_\alpha \in S$. We show that + +$$ + B = \Set{f_\alpha | 0 < \alpha \leq\kappa, \mathbf{f}_\alpha (\alpha) \neq 0} +$$ + +is a basis for $S$. First, suppose that $\sum_{i=1}^n \rho_i f_{\alpha_i} = 0$ where $\alpha_i < \alpha_j$ for $i < j$. Applying this to $\alpha_n$ gives $\rho_n f_{\alpha_n} (\alpha_n) = 0$, and since $R$ is an integral domain, $\rho_n = 0$. Similarly $\rho_i = 0$ for all $i$ and so $B$ is linearly independent. + +To show that $B$ spans $B$, since any $\mathbf{f}\in S$ has finite support, there is a largest index $\alpha_\mathbf{f} = i(\mathbf{f})$ for which $\mathbf{f}(\alpha_\mathbf{f}) \neq 0$. If $\langle B \rangle < S$, then since $\kappa$ is well-ordered, we may choose a $\mathbf{g}\in S\setminus \langle B \rangle$ for which $\alpha = \alpha_\mathbf{g} = i(\mathbf{g})$ is as small as possible. Then $\mathbf{g}\in M$. Moreover, since $0 \neq \mathbf{g}(\alpha)\in\mathcal{I}_\alpha$, it follows that $\mathbf{f}_\alpha (\alpha) \neq 0$ and $\mathbf{g}(\alpha) = c\mathbf{f}_\alpha (\alpha)$ for some $c\in R$. Then $\operatorname{supp}(\mathbf{g} - c\mathbf{f}_\alpha) \subseteq [0, \alpha]$ and + +$$ + (\mathbf{g} - c\mathbf{f}_\alpha)(\alpha) = \mathbf{g}(\alpha) - c\mathbf{f}_\alpha(\alpha) = 0 +$$ + +and so $i(\mathbf{g} - c\mathbf{f}_\alpha) < \alpha$, which implies that $\mathbf{g} - c\mathbf{f}_\alpha \in \langle B \rangle$. But then + +$$ + \mathbf{g} = (\mathbf{g} - c\mathbf{f}_\alpha) + c\mathbf{f}_\alpha \in \langle B \rangle +$$ + +which is a contradiction. Hence, $B$ is a basis for $S$. +
+ + + +Let $M$ be a free $R$-module of finite rank $n$, where $R$ is a principal ideal domain. Let $S = \Set{\mathbf{s}_i}_{i=1}^n$ be a spanning set for $M$. Then $S$ is a bsis for $M$. + +
+Proof + +Let $B = Set{\mathbf{b}_i}_{i=1}^n$ be a basis for $M$ and define the map $f: M\to M$ by $f\mathbf{b} = \mathbf{s}_i$ and extending to a surjective $R$-homomorphism. Since $M$ is free, a previous theorem implies that + +$$ + M \cong \ker(f) \boxplus \operatname{ran}(f) = \ker(f) \boxplus M +$$ + +Since $\ker(f)$ is a submodule of the free module and since $R$ is a principal ideal domain, we know that $\ker(f)$ is free of rank at most $n$. It follows that + +$$ + \operatorname{rank}(M) = \operatorname{rank}(\ker(f)) + \operatorname{rank}(M) +$$ + +and so $\operatorname{rank}(\ker(f)) = 0$, i.e. $\ker(f) = \Set{0}$, which implies that $f$ is an $R$-isomorphism and so $S$ is a basis. +
+ + + +Let $M$ be a free $R$-module of rank $n$, where $R$ is a principal ideal domain. Let $N$ be a submodule of $M$ that is free of rank $k\leq n$. Then there is a basis $B$ for $M$ that contains a subset $S=\Set{\mathbf{v}_i}_{i=1}^k$ for which $\Set{\rho_i \mathbf{v}_i}_{i=1}^k$ is a basis for $N$ for some nonzero elements $\rho_i \in R$. + + +## Torsion-free and free modules + + +A finitely generated module over a principal ideal domain is free if and only if it is torsion-free. + +
+Proof + +It is straightforward to see that a finitely generated module $M$ over a principal ideal domain that is free, it is also torsion-free. Conversely, suppose $M$ is torsion free and let $G = \Set{\mathbf{v}_i}_{i=1}^n$ be a generating set for $M$. In the case $n = 1$, we have $G = \Set{\mathbf{v}}$. Then $G$ is a basis for $M$ since singleton are linearly independent in a torsion-free module. Hence $M$ is free. + +Suppose that $G = \Set{\mathbf{u}, \mathbf{v}}$ is a generating set with $\mathbf{u},\mathbf{v}\neq 0$. If $G$ is linearly independent, we are done. If not, then there exist nonzero $\alpha, \beta \in R$ for which $\alpha\mathbf{u} = \beta\mathbf{v}$. It follows that $\beta M = \beta\langle\mathbf{u},\mathbf{v}\rangle \subseteq \langle\mathbf{u}\rangle$ and so $\beta M$ is a submodule of a free module and is therefore free by the previous proposition. However, the map $f:M\to\beta M$ defined by $f\mathbf{v} = \beta\mathbf{v}$ is an isomorphism because $M$ is torsion-free. Hence, $M$ is also free. + +For the general case, write + +$$ + G = \Set{\mathbf{u}_1,\dots,\mathbf{u}_k,\mathbf{v}_1,\dots,\mathbf{v}_{n-k}} +$$ + +where $S = \Set{\mathbf{u}_i}_{i=1}^k$ is a maximal linearly independent subset of $G$. Note that $S$ is nonempty because singleton sets are linearly independent. + +For each $\mathbf{v}_i$, the set $\Set{\mathbf{u}_i,\dots,\mathbf{u}_k,\mathbf{v}_i}$ is linearly dependent and so there exist $\alpha_i \in R$ and $\rho_1,\dots,\rho_k \in R$ for which + +$$ + \alpha_i \mathbf{v}_i + \sum_{j=1}^k \rho_j \mathbf{u}_j = 0 +$$ + +If $\alpha = \prod_{i=1}^{n-k} \alpha_i$, then + +$$ + \alpha M = \alpha\langle \mathbf{u}_1,\dots,\mathbf{u}_k,\mathbf{v}_1,\dots,\mathbf{v}_{n-k}\rangle \subseteq \langle\mathbf{u}_1,\dots,\mathbf{u}_k \rangle +$$ + +Since the latter is a free module, so is $\alpha M$, and therefore so is $M$. +
+ + +## The primary cyclic decomposition theorem + + +Any finitely generated module $M$ over a principal ideal domain $R$ is the direct sum of a finitely generated free $R$-module and a finitely generated torsion $R$ module + +$$ + M = M_\mathrm{free} \oplus M_\mathrm{tor} +$$ + +The torsion part $M_\mathrm{tor}$ is unique, since it must be the set all torsion elements of $M$, whereas the free part $M_\mathrm{free}$ is unique only up to isomorphism, i.e. the rank of the free part is unique. + +
+Proof + +It is easy to see that the set $M_\mathrm{tor}$ of all torsion elements is a submodule of $M$ and the quotient $M\setminus M_\mathrm{tor}$ is torsion-free. Moreover, since $M$ is finitely generated, so is $M\setminus M_\mathrm{tor}$. Hence, the previous theorem implies that $M\setminus M_\mathrm{tor}$ is free. Hence, by a previous theorem + +$$ + M = M_\mathrm{tor} \oplus F +$$ + +where $F \cong M\setminus M_\mathrm{tor}$ is free. + +As to the uniqueness of the torsion part, suppose that $M = T \oplus G$ where $T$ is torsion and $G$ is free. Then $T \subseteq M_\mathrm{tor}$. However, if $\mathbf{v} = \mathbf{t} + \mathbf{g} \in M_\mathrm{tor}$ for $\mathbf{t}\in T$ and $\mathbf{g}\in G$, then $\mathbf{g} = \mathbf{v} - \mathbf{t}\in M_\mathrm{tor}$ and so $\mathbf{g} = 0$ and $\mathbf{v}\in T$. Thus, $T = M_\mathrm{tor}$. + +For the free part, since $M = M_\mathrm{tor} \oplus F = M_\mathrm{tor} \oplus G$, the submodules $F$ and $G$ are both complements of $M_\mathrm{tor}$ and hence are isomorphic. +
+ + +Note that if $\Set{\mathbf{w}_i}_{i=1}^m$ is a basis for $M_\mathrm{free}$ we can write $M = \left(\bigoplus_{i=1}^m \langle \mathbf{w}_i \rangle\right) M_\mathrm{tor}$, where each cyclic submodule $\langle \mathbf{w}_i \rangle$ has zero annihilator. This is a partial decomposition of $M$ into a direct sum of cyclic submodules. + +### Primary decomposition + + +Let $p$ be a prime in $R$. A $p$-*primary* module is a module whose order is a power of $p$. + + + +Let $M$ be a torsion module over a principal ideal domain $R$, with order + +$$ + \mu = \prod_{i=1}^n p_i^{e_i} +$$ + +where the $p_i$ are distinct nonassociate primes in $R$. +1. $M$ is the direct sum $M = \bigoplus_{i=1}^n M_{p_i}$ where +$$ + M_{p_i} = \frac{\mu}{p_i^{e_i}} M = \Set{\mathbf{v}\in M | p_i^{e_i}\mathbf{v} = 0} +$$ +is a primary submodule of order $p_i^{e_i}$. This decomposition of $M$ into primary submodules is called the *primary decomposition* of $M$. +2. The primary decomposition of $M$ is unique up to order of the summands, i.e. if $M = \bigoplus_{i=1}^m N_{q_i}$, where $N_{q_i}$ is primary of order $q_i^{f_i}$ and $q_1,\dots,q_m$ are distinct nonassociate primes, then $m = n$ and $N_{q_i} = M_{p_i}$ (after reindexing if necessary). Hence $f_i = e_i$ and $q_i \sim p_i$ for $i = 1,\dots,n$. +3. Two $R$-modules $M$ and $N$ are isomorphic if and only if the summands in their primary decomposition are pairwise isomorphic, i.e. if $M = \bigoplus_{i=1}^n M_{p_i}$ and $N = \bigoplus_{i=1}^m N_{q_i}$ are primary decompositions, then $m = n$ and $M_{p_i} \cong N_{q_i}$ (after reindexing if necessary) for $i=1,\dots,n$. + +
+Proof + +Let us write $\mu_i = \frac{\mu}{p_i^{e_i}}$ and show first that + +$$ + M_{p_i} = \mu_i M = \Set{\mu_i \mathbf{v} | \mathbf{v}\in M} +$$ + +Since $p_i^{e_i} (\mu_i M) = \mu M = \Set{0}$, we have $\mu_i M \subseteq M_{p_i}$. On the other hand, since $\mu_i$ and $p_i^{e_i}$ are relatively prime, there exists $a, b\in R$ for which + +$$ + a\mu_i + bp_i^{e_i} = 1 +$$ + +and so if $\mathbf{x}\in M_{p_i}$ then + +$$ + \mathbf{x} = (a\mu_i + bp_i^{e_i})\mathbf{x} = a\mu_i \mathbf{x} \in \mu_i M +$$ + +Hence $M_{p_i} = \mu_i M$. + +**(1):** Since $\gcd(\mu_1,\dots,\mu_n) = 1$, there exist scalars $a_i \in R$ for which $\sum_{i=1}^n a_i \mu_i = 1$, and so for any $\mathbf{x}\in M$ + +$$ + \mathbf{x} = \left(\sum_{i=1}^n a_i \mu_i\right)\mathbf{x} \in \sum_{i=1}^n \mu_i M +$$ + +Moreover, since the $o(\mu_i M) | p_i^{e_i}$ and the $p_i^{e_i}$ are pairwise relatively prime, it follows that the sum of the submodules $\mu_i M$ is direct, i.e. + +$$ + M = \bigoplus_{i=1}^n \mu_i M = \bigoplus_{i=1}^n M_{p_i} +$$ + +As to the annihilators, it is clear that $\langle p_i^{e_i} \subseteq \operatorname{ann}(\mu_i M)$. For the reverse inclusion, if $r\in\operatorname{ann}(\mu_iM)$, then $r\mu_i \in\operatorname{ann}(M)$ and so $p_i^{e_i} \mu_i | r\mu_i$, i.e. $p_i^{e_i} | r$ and so $r\in \langle p_i^{e_i} \rangle$. Hence $\operatorname{ann}(\mu_i M) = \langle p_i^{e_i} \rangle$. + +**(2):** As to uniqueness, we claim that $q = \prod_{i=1}^m q_i^{f_i}$ is an order of $M$. It is clear that $q$ annihilates $M$ and so $\mu|q$. On the other hand, $N_{q_i}$ contains an element $u_i$ of order $q_i^{f_i}$, and so the sum $\mathbf{v} = \sum_{i=1}^m \mathbf{u}_i$ has order $q$, which implies that $q | \mu$. Hence, $q$ and $\mu$ are associates. + +Unique factorization in $R$ implies that $m = n$ and, after a suitable reindexing, that $f_i = e_i$ and $q_i$ and $p_i$ are associates. Hence $N_{q_i}$ is primary of order $p_i^{e_i}$. For convenience, we can write $N_{q_i}$ as $N_{p_i}$. Hence + +$$ + N_{p_i} \subseteq \Set{\mathbf{v} \in M | p_i^{e_i} \mathbf{v} = 0} = M_{p_i} +$$ + +However, if $\bigoplus_{i=1}^n N_{p_i} = \bigoplus_{i=1}^n M_{p_i}$ and $N_{p_i} \subseteq M_{p_i}$, we must have $N_{p_i} = M_{p_i}$ for all $i$. + +**(3):** If $m = n$ and $g_i: M_{p_i} \cong N_{q_i}$, then the map $g: M\to N$ defined by $h\left(\sum_{i=1}^n \mathbf{a}_i \right) = \sum_{i=1}^n g_i (\mathbf{a}_i)$ is an isomorphism and so $M \cong N$. Conversely, suppose that $g: M\cong N$. Then $M$ and $N$ have the same annihilators and therefore the same order + +$$ + \mu = \prod_{i=1}^n p_i^{e_i} +$$ + +Hence, **(1)** and **(2)** imply that $m = n$ and after a suitable reindexing, $q_i = p_i$. Moreover, since + +$$ + \mathbf{a}\in M_{p_i} \iff \mu_i \mathbf{a} = 0 \iff g(\mu_i\mathbf{a}) = 0 \iff \mu_i g\mathbf{a} = 0 \iff g\mathbf{a} \in N_{p_i} +$$ + +it follows that $g: M_{p_i}\cong N_{p_i}$. +
+ + +### Cyclic decomposition of primary modules + + +Let $M$ be a module over a principal ideal domain $R$ and let $p\in R$ be a prime. +1. If $pM = \Set{0}$, then $M$ is a vector space over the field $R\setminus\langle p \rangle$ with scalar multiplication defined by $(\rho + \langle p\rangle)\mathbf{v} = \rho\mathbf{v}$ for all $\mathbf{v}\in M$ and $\rho\in R$. +2. For any submodules $S$ of $M$, the set $S^{(p)} = \Set{\mathbf{v}\in S | p\mathbf{v} = 0}$ is also a submodule of $M$ and if $M = S\oplus T$, then $M^{(p)} = S^{(p)} \oplus T^{(p)}$. + +
+Proof + +**(1):** Since $p$ is a prime, the ideal $\langle p \rangle$ is maximal and so $R\setminus \langle p \rangle$ is a field. + +**(2):** Since $S^{(p)} \subseteq S$ and $T^{(p)} \subseteq T$ we see that $S^{(p)} \cap T^{(p)} = \Set{0}$. Also, if $\mathbf{v}\in M^{(p)}$, then $p\mathbf{v} = 0$. However, $\mathbf{v} = \mathbf{s} + \mathbf{t}$ for some $\mathbf{s}\in S$ and $\mathbf{t}\in T$ and so $0 = p\mathbf{v} = p\mathbf{s} + p\mathbf{t}$. Since $p\mathbf{s}\in S$ and $p\mathbf{t}\in T$, we deduce that $p\mathbf{s} = p\mathbf{t} = 0$, such that $\mathbf{v}\in S^{(p)} \oplus T^{(p)}$. Thus, $M^{(p)} \subseteq S^{(p)} \oplus T^{(p)}$. The reverse inclusion is manifest. +
+ + + +Let $M$ be a primary finitely generated torsion module over a principal ideal domain $R$, with order $p^e$. +1. $M$ is a direct sum $M = \bigoplus_{i=1}^n \langle \mathbf{v}_i \rangle$ of cyclic submodules with annihilators $\operatorname{ann}(\langle \mathbf{v}_i \rangle) = \langle p^{e_i} \rangle$, which can be arranged in ascending order +$$ + \operatorname{ann}(\langle \mathbf{v}_1 \rangle) \subseteq\cdots\subseteq \operatorname{ann}(\langle \mathbf{v}_n \rangle) +$$ +or equivalently, $e = e_1 \geq\cdots\geq e_n$ +2. If $M$ is also the direct sum $M = \bigoplus_{i=1}^n \langle \mathbf{u}_i \rangle$ of cyclic submodules with annihilators $\operatorname{ann}(\langle\mathbf{u}_i\rangle) = \langle q^{f_i} \rangle$, arranged in ascending order +$$ + \operatorname{ann}(\langle \mathbf{u}_1 \rangle) \subseteq\cdots\subseteq \operatorname{ann}(\langle \mathbf{u}_n \rangle) +$$ +or equivalently $f_1 \geq\cdots\geq f_m$. Then the two chains of annihilators are identical, i.e. $m = n$ and +$$ + \operatorname{ann}(\langle \mathbf{u}_i \rangle) = \operatorname{ann}(\langle \mathbf{v}_i \rangle),\;\forall $i$ +$$ +Thus, $p\sim q$ and $f_i = e_i$ for all $i$. +3. Two $p$-primary $R$-modules $M = \bigoplus_{i=1}^n \langle \mathbf{v}_i \rangle$ and $N = \bigoplus_{i=1}^n \langle \mathbf{u}_i \rangle$ are isomorphic if and only if they have the same annihilator chains, i.e. if and only if $m = n$ and, after a possible reindexing +$$ + \operatorname{ann}(\langle \mathbf{u}_i \rangle) = \operatorname{ann}(\langle \mathbf{v}_i \rangle),\;\forall $i$ +$$ + +
+Proof + +**(1):** Let $\mathbf{v}_1 \in M$ have order equal to the order of $M$, i.e. $\operatorname{ann}(\mathbf{v}_1) = \operatorname{ann}(M) = \langle p^e \rangle$. Such an element must exist since $o(\mathbf{v}_1) \leq p^e$ for all $\mathbf{v}\in M$ and if this inequality is strict, then $p^{e-1}$ will annihilate $M$. + +If we show that $\langle \mathbf{v}_1 \rangle$ is complemented, i.e. $M = \langle\mathbf{v}_1 \rangle \oplus S_1$ for some submodule $S_1$, then since $S_1$ is also a finitely generated primary torsion module over $R$, we can repeat the process to get + +$$ + M = \langle\mathbf{v}_1 \rangle \oplus \langle\mathbf{v}_2 \rangle \oplus S_2 +$$ + +where $\operatorname{ann}(\mathbf{v}_i) = \langle p^{e_i} \rangle$. We can continue this decomposition + +$$ + M = \left( \oplus_{i=1} \langle \mathbf{v}_1 \rangle \right) \oplus S_n +$$ + +as long as $S_n \neq \Set{0}$. However, the ascending sequences of submodules +$$ + \langle\mathbf{v}_1 \rangle \subseteq \langle \mathbf{v}_1 \rangle \oplus \langle \mathbf{v}_2 \rangle \subseteq\cdots +$$ + +must terminate since $M$ is Noetherian and so there is an integer $n$ for which eventually $S_n = \Set{0}$ giving $M = \bigoplus_{i=1}^n \langle \mathbf{v}_i \rangle$. + +**(2):** Let $\mathbf{v} = \mathbf{v}_1$. The direct sum $M_1 = \langle \mathbf{v} \rangle \oplus\Set{0}$ clearly exists. Suppose that the direct sum $M_k = \langle \mathbf{v} \rangle \oplus S_k$ exists. We claim that if $M_m \leq M$, the it is possible to find a submodule $S_{k+1}$ for which $S_k < S_{k+1}$ and for which the direct sum $M_{k+1} = \langle \mathbf{v} \rangle \oplus S_{k+1}$ also exists. This process must also stop after a finite number of steps, giving $M = \langle \mathbf{v} \rangle \oplus S$. + +If $M_k < M$ and $\mathbf{u}\in M\setminus M_1$, let $S_{k+1} = \langle S_k , \mathbf{u} - \alpha\mathbf{v}\rangle$ for $\alpha\in R$. Then $S_k < S_{k+1}$ since $\mathbf{u}\notin M_k$. We want to show that for some $\alpha\in R$, the direct sum $\langle\mathbf{v}\rangle\oplus S_{k+1}$ exists, i.e. + +$$ + \mathbf{x} \in \langle \mathbf{v} \rangle \cap \langle S_k, \mathbf{u} - \alpha\mathbf{v} \rangle \implies \mathbf{x} = 0 +$$ + +Now, there exists scalars $a$ and $b$ for which $\mathbf{x} = a\mathbf{v} = \mathbf{s} + b(\mathbf{u} - \alpha\mathbf{v})$ for $\mathbf{s}\in S_k$ and so if we find a scalar $\alpha$ for which $b(\mathbf{u} - \alpha\mathbf{v})\in S_k$ then $\langle \mathbf{v} \rangle \cap S_k = \Set{0}$ implies that $\mathbf{x} = 0$, completing the proof of existence. + +Solving for $b\mathbf{u}$ gives + +$$ + b\mathbf{u} = (a + \alpha b)\mathbf{v} - \mathbf{s} \in \langle\mathbf{v} \rangle \oplus S_k = M_k +$$ + +so let us consider the ideal of all such scalars $\mathcal{I} = \Set{\rho\in R | \rho\mathbf{u} \in M_k}$. Since $p^e \in\mathcal{I}$ and $\mathcal{I}$ is principal, we have $\mathcal{I} = \langle p^f \rangle$ for some $f \leq e$. Also, $f > 0$ since $\mathbf{u}\notin M_k$ implies that $1\notin \mathcal{I}$. Since $b\in\mathcal{I}$, we have $b = \beta p^f$ and there exist $d\in R$ and $\mathbf{t}\in S_k$ for which $p^f \mathbf{u} = d\mathbf{v} + \mathbf{t}$. Hence + +$$ + b\mathbf{u} = \beta p^f \mathbf{u} = \beta(d\mathbf{v} + \mathbf{t}) = \beta d\mathbf{v} + \beta\mathbf{t} +$$ + +Now we need more information about $d$. Multiplying the expression for $p^f \mathbf{u}$ by $p^{e-f}$ gives + +$$ + 0 = p^{e}\mathbf{u} = p^{e-f}(p^f \mathbf{u}) = p^{e-f} d\mathbf{v} + p^{e-f}\mathbf{t} +$$ + +and since $\langle\mathbf{v}\rangle \cap S_k = \Set{0}$, it follows that $p^{e-f} d\mathbf{v} = 0$. Hence, $p^e | p^{e-f} d$, i.e. $p^f | d$ and so $d = \delta p^f$ for some $\delta\in R$. Now we can write + +$$ + b\mathbf{u} = \beta\delta p^f \mathbf{v} + \beta\mathbf{t} +$$ + +and so + +$$ + b(\mathbf{u} - \delta\mathbf{v}) = \beta\mathbf{t} \in S_k +$$ + +Thus, we take $\alpha = \delta$ to get $b(\mathbf{u} - \alpha\mathbf{v}) \in S_k$, completing the proof of existence. + +For uniqueness, note first that $M$ has orders $p^{e_1}$ and $q^{f_1}$ and so $p$ and $q$ are associates and $e_1 = f_1$. Next, we show that $n = m$. According to **(2)** of the previous lemma $M^{(p)} = \bigoplus_{i=1} \langle \mathbf{v}_i \rangle^{(p)}$ and $M^{(p)} = \bigoplus_{i=1} \langle \mathbf{u}_i \rangle^{(p)}$ where all the summands are nonzero. Since $pM^{(p)} = \Set{0}, it follows from the previous lemma that $M^{(p)}$ is a vector space over $R\setminus\langle p \rnagle$ and so each of the preceding decompositions expresses $M^(p)$ as a direct sim of one-dimensional vector subspaces. Hence $m = \dim(M^{(p)}) = n$. + +Finally, we show that the exponents $e_i$ and $f_i$ are equal using induction on $e_1$. If $e_1 = 1$, then $e_i = 1$ for all $i$ and since $f_1 = e_1$, we also have $f_i = 1$ for all $i$. Suppose that the result is true whenever $e_1 \leq k-1$ and let $e_1 = k$. Write + +$$ +\begin{align*} + (e_1,\dots,e_n) =& (e_1,\dots,e_n,1,\dots,1),\; e_s > 1 \\ + (f_1,\dots,f_n) =& (f_1,\dots,f_t,1,\dots,1),\; f_t > 1 +\end{align*} +$$ + +Then $pM = \bigoplus_{i=1}^s p\langle\mathbf{v}_i \rangle$ and $pM = \bigoplus_{i=1}^t p\langle\mathbf{u}_i \rangle$. However, $p\langle \mathbf{v}_1 \rangle = \langle p\mathbf{v}_1 \rangle$ is a cyclic submodule of $M$ with annihilator $\langle p^{e_i - 1}\rangle$ and so by the induction hypothesis $\mathbf{s} = \mathbf{t}$ and $e_1 = f_1,\dots,e_s = f_s$, concluding the proof of uniqueness. + +**(3):** Suppose $g:M\cong N$ and $M$ has annihilator chain + +$$ + \operatorname{ann}(\langle\mathbf{v}_1\rangle) \subseteq\cdots\subseteq \operatorname{ann}(\langle\mathbf{v}_n \rangle) +$$ + +and $N$ has annihilator chain + +$$ + \operatorname{ann}(\langle\mathbf{u}_1\rangle) \subseteq\cdots\subseteq \operatorname{ann}(\langle\mathbf{u}_n \rangle) +$$ + +Then + +$$ + N = gM = \bigoplus_{i=1} \langle g\mathbf{v}_1 \rangle +$$ + +and so $m = n$ and after a suitable reindexing + +$$ + \operatorname{ann}(\langle\mathbf{v}_i\rangle) = \operatorname{ann}(\langle g\mathbf{v}_i \rangle) = \operatorname{ann}(\langle\mathbf{u}_i \rangle) +$$ + +Conversely, suppose $M = \bigoplus_{i=1}^n \mathbf{v}_i$ and $N = \bigoplus_{i=1}^n \langle \mathbf{u}_i \rangle$ have the same annihilator chains, i.e. $m = n$ and + +$$ + \operatorname{ann}(\langle\mathbf{u}_i\rangle) = \operatorname{ann}(\langle\mathbf{v}_i \rangle) +$$ + +Then + +$$ + \langle\mathbf{u}_i \rangle \cong \frac{R}{\operatorname{ann}(\langle\mathbf{u}_i \rangle)} = \frac{R}{\operatorname{ann}(\langle\mathbf{v}_i \rangle)} \cong \langle\mathbf{v}_i \rangle +$$ +
+ + +### Primary cyclic decomposition + + +Let $M$ be a finitely generated torsion module over a principal ideal domain $R$. +1. If $M$ has order $\mu = \prod_{i=1}^n p_i^{e_i}$ where the $p_i$ are distinct nonassociate primes in $R$, then $M$ can be uniquely decomposed (up to the order of the summands) into the direct sum $M = \bigoplus_{i=1} M_{p_i}$ where +$$ + M_{p_i} = \frac{\mu}{p_i^{e_i}} M = \Set{\mathbf{v}\in M | p_i^{e_i}\mathbf{v} = 0} +$$ +is a primary submodule with annihilator $\langle p_i^{e_i} \rangle$. Finally, each primary submodule $M_{p_i}$ can be written as a direct sum of cyclic submodules, so that +$$ + M = \underbrace{[\langle \mathbf{v}_{1,1}\oplus\cdots\oplus\langle \mathbf{v}_{i, k_1} \rangle]}_{M_{p_1}} \oplus\cdots\oplus \underbrace{[\langle \mathbf{v}_{1,1}\oplus\cdots\oplus\langle \mathbf{v}_{i, k_1} \rangle]}_{M_{p_n}} +$$ +where $\operatorname{ann}(\langle\mathbf{v}_{i,j}\rangle) = \langle p_i^{e_{i,j}}\rangle$ and the terms in each cyclic decomposition can be arranged so that, for each $i$ +$$ + \operatorname{ann}(\langle\mathbf{v}_{i,1}) \subseteq\cdots\subseteq \operatorname{ann}(\langle\mathbf{v}_{i,k}\rangle) +$$ +or equivalently $e_i = e_{i,1} \geq\cdots\geq e_{i,k_i}$. +2. As for uniqueness, suppose +$$ + M = \underbrace{[\langle \mathbf{u}_{1,1}\oplus\cdots\oplus\langle \mathbf{u}_{i, k_1} \rangle]}_{N_{q_1}} \oplus\cdots\oplus \underbrace{[\langle \mathbf{u}_{1,1}\oplus\cdots\oplus\langle \mathbf{u}_{i, k_1} \rangle]}_{N_{q_n}} +$$ +is also a primary cyclic decomposition of $M$. Then + - The number of summands is the same in both decompositions; in fact, $m = n$ and after possible reindexing $k_u = j_u$ for all $u$. + - The primary submodules are the same, i.e. after possible reindexing $q_i \sim p_i$ and $N_{q_i} = M_{p_i}$. + - For each primary submodule pair $N_{q_i} = M_{p_i}$, the cyclic submodules have the same annihilator chains, i.e. after possible reindexing $\operatorname{ann}(\langle \mathbf{u}_{i,j} \rangle) = \operatorname{ann}(\langle \mathbf{v}_{i,j} \rangle)$ for all $i,j$. +In summary, the primary submodules and annihilator chains are uniquely determined by the module $M$. +3. Two $R$-modules $M$ and $N$ are isomorphic if and only if they have the same annihilator chains. + + +### Elementary divisors + +## Invariant factor decomposition + +## Characterizing cyclic modules + +## Indecomposable modules + +### Indecomposable submodules of prime order + # Linear operators ## Eigenvalues and eigenvectors diff --git a/content/notes/math/manifolds.mdx b/content/notes/math/manifolds.mdx index ad936e5..f43cc25 100644 --- a/content/notes/math/manifolds.mdx +++ b/content/notes/math/manifolds.mdx @@ -1129,8 +1129,8 @@ Suppose $M$ and $N$ are smooth manifolds with or without boundary, and that $F:M Suppose $M$ is a topological space, and let $\mathcal{X} = \Set{X_\alpha}_{\alpha\in A}$ be an arbitrary open cover of $M$, indexed by a set $A$. A *partition of unity* subordinate to $\mathcal{X}$ is an indexed collection of continuous function $\Set{\psi_\alpha:M\to\R}_{\alpha\in A}$ with the following properties: 1. $0\leq\psi_\alpha(x)\leq 1$ for all $\alpha\in A$ and all $x\in M$ -2. $\mathrm{supp}(\psi_\alpha) = \overline{\Set{p\in M | \psi_\alpha(p) \neq 0}}\subseteq X_\alpha$ for each $\alpha\in A$. -3. The collection of supports $\Set{\mathrm{supp}(\psi_\alpha)}_{\alpha\in A}$ is locally finite, meaning that every point has a neighbourhood that intersects $\mathrm{supp}(\psi_\alpha)$ for only finitely many values of $\alpha$. +2. $\operatorname{supp}(\psi_\alpha) = \overline{\Set{p\in M | \psi_\alpha(p) \neq 0}}\subseteq X_\alpha$ for each $\alpha\in A$. +3. The collection of supports $\Set{\operatorname{supp}(\psi_\alpha)}_{\alpha\in A}$ is locally finite, meaning that every point has a neighbourhood that intersects $\operatorname{supp}(\psi_\alpha)$ for only finitely many values of $\alpha$. 4. $\sum_{\alpha\in A} \psi_\alpha (x) = 1$ for all $x\in M$ If $M$ is a smooth manifold with or without boundary, a *smooth parition of unity* is one for which each of the functions $\psi_\alpha$ is smooth. @@ -1153,7 +1153,7 @@ $$ \end{cases} $$ -where $H_i:\R^n \to\R$ is a smooth function that is positive in $B_{r_i}(\mathbf{0})$ and zero elsewhere. On the set $B'_i\setminus\overline{B}_i$ where the two definitions overlap, both definitions yield the zero function, $f_i$ is well-defined and smooth with $\mathrm{supp}(f_i) = \overline{B}_i$. +where $H_i:\R^n \to\R$ is a smooth function that is positive in $B_{r_i}(\mathbf{0})$ and zero elsewhere. On the set $B'_i\setminus\overline{B}_i$ where the two definitions overlap, both definitions yield the zero function, $f_i$ is well-defined and smooth with $\operatorname{supp}(f_i) = \overline{B}_i$. Define $f:M\to\R$ by $f(x) = \sum_i f_i(x)$. Because of the local finiteness of the cover $\Set{\overline{B}_i}$, this sum has only finitely many nonzero terms in a neighbourhood of each point and thus defines a smooth function. Because each $f_i$ is nonnegative everywhere and positive on $B_i$, and every point of $M$ is in some $B_i$, it follows that $f(x) > 0$ everywhere on $M$. Thus, the functions $g_i:M\to\R$ defined by $g_i (x) \frac{f_i (x)}{f(x)}$ are also smooth. It is immediate from the definition that $0\leq g_i \leq 1$ and $\sum_i g_i \equiv 1$. @@ -1166,17 +1166,17 @@ $$ If there are no indices $i$ for which $a(i) = \alpha$, then this sum should be interpreted as the zero function. It follos that $$ - \mathrm{supp}(\psi_\alpha) = \overline_{\bigcup_{i:a(i) = \alpha} B_i} = \bigcup_{i:a(i) = \alpha} \overline{B}_i \subseteq X_\alpha + \operatorname{supp}(\psi_\alpha) = \overline_{\bigcup_{i:a(i) = \alpha} B_i} = \bigcup_{i:a(i) = \alpha} \overline{B}_i \subseteq X_\alpha $$ -Each $\psi_\alpha$ is a smooth function that satisfies $0\leq\psi_\alpha \leq 1$. Moreover, the collection of supports $\Set{\mathrm{supp}(\psi_\alpha)}_{\alpha\in A}$ is still locally finite, and $\sum_\alpha \psi_\alpha \equiv \sum_i g_i \equiv 1$, which is the desired partition of unity. +Each $\psi_\alpha$ is a smooth function that satisfies $0\leq\psi_\alpha \leq 1$. Moreover, the collection of supports $\Set{\operatorname{supp}(\psi_\alpha)}_{\alpha\in A}$ is still locally finite, and $\sum_\alpha \psi_\alpha \equiv \sum_i g_i \equiv 1$, which is the desired partition of unity. #### Applications of partitions of unity -If $M$ is a topological space, $A\subseteq M$ is a closed subset, and $U\subseteq M$ is an open subset containing $A$, a continuous function $\psi:M\to\R$ is called a *bump function* for $A$ supported in $U$ if $0\leq\psi\leq 1$ on $M$ and $\psi\equiv 1$ on $A$ and $\mathrm{supp}(\psi) \subseteq U$. +If $M$ is a topological space, $A\subseteq M$ is a closed subset, and $U\subseteq M$ is an open subset containing $A$, a continuous function $\psi:M\to\R$ is called a *bump function* for $A$ supported in $U$ if $0\leq\psi\leq 1$ on $M$ and $\psi\equiv 1$ on $A$ and $\operatorname{supp}(\psi) \subseteq U$. @@ -1190,14 +1190,14 @@ Let $U_0 = U$ an $U_1 = M\setminus A$ and let $\Set{\psi_0, \psi_1}$ be a smooth -Suppose $M$ is a smooth manifold with or without boundary, $A\subseteq M$ is a closed subset, and $f:A\to\R^k$ is a smooth function. For any open subset $U$ containing $A$, there exists a smooth function $\tilde{f}:M\to\R^k$ such that $\tilde{f}|_A = f$ and $\mathrm{supp}(\tilde{f})\subseteq U$. +Suppose $M$ is a smooth manifold with or without boundary, $A\subseteq M$ is a closed subset, and $f:A\to\R^k$ is a smooth function. For any open subset $U$ containing $A$, there exists a smooth function $\tilde{f}:M\to\R^k$ such that $\tilde{f}|_A = f$ and $\operatorname{supp}(\tilde{f})\subseteq U$.
Proof -For each $p\in A$, choose a neighbourhood $W_p$ of $p$ and a smooth function $\tilde{f}_p :W_p \to R^k$ that agrees with $f$ on $W_p \cap A$. Replacing $W_p$ by $W_p \cap U$, we may assume $W_p \subseteq U$. The collection of sets $\Set{W_p | p\in A}\cup M\setminus A$ is an open cover of $M$. Let $\Set{\psi_p | p\in A} \cup \Set{\psi_0}$ be a smooth partition of unity subordinate to this cover, with $\mathrm{supp}(\psi_p)\subseteq W_p$ and $\mathrm{supp}(\psi_0) \subseteq M\setminus A$. +For each $p\in A$, choose a neighbourhood $W_p$ of $p$ and a smooth function $\tilde{f}_p :W_p \to R^k$ that agrees with $f$ on $W_p \cap A$. Replacing $W_p$ by $W_p \cap U$, we may assume $W_p \subseteq U$. The collection of sets $\Set{W_p | p\in A}\cup M\setminus A$ is an open cover of $M$. Let $\Set{\psi_p | p\in A} \cup \Set{\psi_0}$ be a smooth partition of unity subordinate to this cover, with $\operatorname{supp}(\psi_p)\subseteq W_p$ and $\operatorname{supp}(\psi_0) \subseteq M\setminus A$. -For each $p\in A$, the product $\psi_p \tilde{f}_p$ is smooth on $W_p$, and has a smooth extension to all of $M$ if we interpret it to be zero on $M\setminus\mathrm{supp}(\psi_p)$. Thus, we can defined $\tilde{f}: M\to\R^k$ by +For each $p\in A$, the product $\psi_p \tilde{f}_p$ is smooth on $W_p$, and has a smooth extension to all of $M$ if we interpret it to be zero on $M\setminus\operatorname{supp}(\psi_p)$. Thus, we can defined $\tilde{f}: M\to\R^k$ by $$ \tilde{f}(x) = \sum_{p\in A} \psi_p (x)\tilde{f}_p (x) @@ -1216,7 +1216,7 @@ $$ so $\tilde{f}$ is indeed an extension of $f$. It follows that $$ - \mathrm{supp}(\tilde{f}) = \overline{\bigcup_{p\in A}\mathrm{supp}(\psi_p)} = \bigcup_{p\in A} \mathrm{supp}(\psi_p) \subseteq U + \operatorname{supp}(\tilde{f}) = \overline{\bigcup_{p\in A}\operatorname{supp}(\psi_p)} = \bigcup_{p\in A} \operatorname{supp}(\psi_p) \subseteq U $$
@@ -2908,7 +2908,7 @@ $$ we see that $\mathbf{v}\in T_\mathbf{p} S$ if and only if $v^i = 0$ for $i > k$. -Let $\varphi$ be a smooth bump function supported in $U$ that is equal to $1$ in a neighbourhood of $\mathbf{p}$. Choose an index $j > k$, and consider the function $f(x) = \varphi(\mathbf{x})x^j$, extended to be zero on $M\setminus\mathrm{supp}(\varphi)$. Then $f$ vanishes identically on $S$, so +Let $\varphi$ be a smooth bump function supported in $U$ that is equal to $1$ in a neighbourhood of $\mathbf{p}$. Choose an index $j > k$, and consider the function $f(x) = \varphi(\mathbf{x})x^j$, extended to be zero on $M\setminus\operatorname{supp}(\varphi)$. Then $f$ vanishes identically on $S$, so $$ 0 = \mathbf{v}f = \sum_{i=1}^n v^i \frac{\partial (\varphi(\mathbf{x}x^j))}{\partial x^i}(\mathbf{p}) = v^i \delta_i^j = v^j @@ -3499,7 +3499,7 @@ $$ The support of $X$ is defined as $$ - \mathrm{supp}(X) = \overline{\Set{\mathbf{p}\in M | X_\mathbf{p} \neq \mathbf{0}}} + \operatorname{supp}(X) = \overline{\Set{\mathbf{p}\in M | X_\mathbf{p} \neq \mathbf{0}}} $$ A vector field is *compactly supported* if its support is a compact set. @@ -3538,7 +3538,7 @@ where $X^i$ is the $i$th component function of $X$ in $x^i$-coordinates. It foll -Let $M$ be a smooth manifold with or without boundary, and let $A\subseteq M$ be a closed subset. Suppose $X$ is a smooth vector field along $A$. Given any subset $U$ containing $A$, there exists a smooth global vector field $\tilde{X}$ on $M$ such that $\tilde{X}|_A$ and $\mathrm{supp}(\tilde{X})\subseteq U$. +Let $M$ be a smooth manifold with or without boundary, and let $A\subseteq M$ be a closed subset. Suppose $X$ is a smooth vector field along $A$. Given any subset $U$ containing $A$, there exists a smooth global vector field $\tilde{X}$ on $M$ such that $\tilde{X}|_A$ and $\operatorname{supp}(\tilde{X})\subseteq U$. @@ -3629,7 +3629,7 @@ $$ Since the component functions $X^i$ are smooth on $U$, if follows that $Xf$ is smooth in $U$. Since the same is true in a neighbourhood of each point, $Xf$ is smooth on $M$. **(2)** $\implies$ **(3)**
-Suppose $U\subseteq M$ is open and $f\in C^\infty (U)$. For any $\mathbf{p}\in U$, let $\psi$ be a smooth bump function that is equal to $1$ in a neighbourhood of $\mathbf{p}$ and supported in $U$, and define $\tilde{f} = \psi f$, extended to be zero on $M\setminus\mathrm{supp}(\psi)$. Then $X\tilde{f}$ is smooth by assumption, and is equal to $Xf$ in a neighbourhood of $\mathbf{p}$. This shows that $Xf$ is smooth in a neighbourhood of each point of $U$. +Suppose $U\subseteq M$ is open and $f\in C^\infty (U)$. For any $\mathbf{p}\in U$, let $\psi$ be a smooth bump function that is equal to $1$ in a neighbourhood of $\mathbf{p}$ and supported in $U$, and define $\tilde{f} = \psi f$, extended to be zero on $M\setminus\operatorname{supp}(\psi)$. Then $X\tilde{f}$ is smooth by assumption, and is equal to $Xf$ in a neighbourhood of $\mathbf{p}$. This shows that $Xf$ is smooth in a neighbourhood of each point of $U$. **(3)** $\implies$ **(1)**
Suppose $Xf$ is smooth whenever $f$ is smooth on an open subset of $M$. If $(x^i)$ are any smooth local coordinates on $U\subseteq M$, we can think of each coordinate $x^i$ as a smooth function on $U$. Applying $X$ to one of these functions, we obtain @@ -4584,7 +4584,7 @@ Every compactly supported smooth vector field on a smooth manifold is complete.
Proof -Suppose $V$ is a compactly supported vector field on a smooth manifold $M$, and let $K = \mathrm{supp}(V)$. For each $\mathbf{p}\in K$, there is a neighbourhood $U_\mathbf{p}$ of $\mathbf{p}$ and a positive number $\varepsilon_\mathbf{p}$ such that the flow of $V$ is defined at least on $(-\varepsilon_\mathbf{p}, \varepsilon_\mathbf{p})\times U_\mathbf{p}$. By compactness, finitely many such sets $U_{\mathbf{p}_i},\dots,U_{\mathbf{p}_k}$ cover $K$. With $\varepsilon = \min\Set{\varepsilon_{\mathbf{p}_1},\dots,\varepsilon_{\mathbf{p}_k}}$, it follows that every maximal integral curve starting in $K$ is defined at least on $(-\varepsilon, \varepsilon)$. Since $V \equiv 0$ outside of $K$, every integral curve starting in $M\setminus K$ is constant and thus can be defined on all of $\R$. Thus the hypotheses of the uniform time lemma are satisfied, so $V$ is complete. +Suppose $V$ is a compactly supported vector field on a smooth manifold $M$, and let $K = \operatorname{supp}(V)$. For each $\mathbf{p}\in K$, there is a neighbourhood $U_\mathbf{p}$ of $\mathbf{p}$ and a positive number $\varepsilon_\mathbf{p}$ such that the flow of $V$ is defined at least on $(-\varepsilon_\mathbf{p}, \varepsilon_\mathbf{p})\times U_\mathbf{p}$. By compactness, finitely many such sets $U_{\mathbf{p}_i},\dots,U_{\mathbf{p}_k}$ cover $K$. With $\varepsilon = \min\Set{\varepsilon_{\mathbf{p}_1},\dots,\varepsilon_{\mathbf{p}_k}}$, it follows that every maximal integral curve starting in $K$ is defined at least on $(-\varepsilon, \varepsilon)$. Since $V \equiv 0$ outside of $K$, every integral curve starting in $M\setminus K$ is constant and thus can be defined on all of $\R$. Thus the hypotheses of the uniform time lemma are satisfied, so $V$ is complete.
@@ -4970,10 +4970,10 @@ $$ On the other hand, $[V,W]_\mathbf{u}$ is easily seen to be equal to the same expression by the Lie bracket formula. -**Case 2: $\mathbf{p}\in\mathrm{supp}(V)$**
-Because $\mathrm{supp}(V)$ is the closure of $R(V)$, it follows by continuity from case 1 that $(\mathcal{L}_V W)_\mathbf{p} = [V,W]_\mathbf{p}$ for $\mathbf{p}\in\mathrm{supp}(V)$. +**Case 2: $\mathbf{p}\in\operatorname{supp}(V)$**
+Because $\operatorname{supp}(V)$ is the closure of $R(V)$, it follows by continuity from case 1 that $(\mathcal{L}_V W)_\mathbf{p} = [V,W]_\mathbf{p}$ for $\mathbf{p}\in\operatorname{supp}(V)$. -**Case 3: $\mathbf{p}\in M\setminus\mathrm{supp}(V)$**
+**Case 3: $\mathbf{p}\in M\setminus\operatorname{supp}(V)$**
In this case, $V \equiv 0$ on a neighbourhood of $\mathbf{p}$. On the one hand, this implies that $\theta_t$ is equal to the identity map in a neighbourhood of $\mathbf{p}$ for all $t$, so $\mathrm{d}(\theta_{-t})_{\theta_t(\mathbf{p})}(W_{\theta_t(\mathbf{p})}) = W_\mathbf{p}$, which implies that $(\mathcal{L}_V W)_\mathbf{p} = 0$. On the other hand $[V, W]_\mathbf{p} = 0$. @@ -5513,7 +5513,7 @@ $$ The *support* of a section $\sigma$ is the set $$ - \mathrm{supp}(\sigma) = \overline{\Set{\mathbf{p}\in M | \sigma(\mathbf{p}) \neq 0}} + \operatorname{supp}(\sigma) = \overline{\Set{\mathbf{p}\in M | \sigma(\mathbf{p}) \neq 0}} $$ If $E\to M$ is a smooth vector bundle, the set of all smooth global sections of $E$ is a vector space under point addition and scalar multiplication @@ -5530,7 +5530,7 @@ $$ -Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold $M$ with or without boundary. Suppose $A$ is a closed subset of $M$, and that $\sigma: A\to E$ is a section of $E|_A$ that is smooth in the sense that $\sigma$ extends to a smooth local section of $E$ is a neighbourhood of each point. For each open subset $U\subseteq M$ containing $A$, there exists a global smooth section $\tilde{\varphi}\in\Gamma(E)$ such that $\tilde{\varphi}|_A = \sigma$ and $\mathrm{supp}(\tilde{\varphi})\subseteq U$. +Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold $M$ with or without boundary. Suppose $A$ is a closed subset of $M$, and that $\sigma: A\to E$ is a section of $E|_A$ that is smooth in the sense that $\sigma$ extends to a smooth local section of $E$ is a neighbourhood of each point. For each open subset $U\subseteq M$ containing $A$, there exists a global smooth section $\tilde{\varphi}\in\Gamma(E)$ such that $\tilde{\varphi}|_A = \sigma$ and $\operatorname{supp}(\tilde{\varphi})\subseteq U$. ### Local and global frames @@ -7605,7 +7605,7 @@ $$ g = \sum_\alpha \psi_\alpha g_\alpha $$ -with each term interpreted to be zero outside $\mathrm{supp}(\mathbf{psi}_\alpha)$. By local finiteness, there are only finitely many nonzero terms in a neighbourhood of each point, so this expression defines a smooth tensor field. It is obviously symmetric, so only positivity needs to be checked. If $\mathbf{v}\in T_\mathbf{p} M$ is any nonzero vector, then +with each term interpreted to be zero outside $\operatorname{supp}(\mathbf{psi}_\alpha)$. By local finiteness, there are only finitely many nonzero terms in a neighbourhood of each point, so this expression defines a smooth tensor field. It is obviously symmetric, so only positivity needs to be checked. If $\mathbf{v}\in T_\mathbf{p} M$ is any nonzero vector, then $$ g_\mathbf{p}(\mathbf{v},\mathbf{v}) = \sum_\alpha \psi_\alpha (\mathbf{p}) g_\alpha |_\mathbf{p} (\mathbf{v}, \mathbf{v}) @@ -9540,7 +9540,7 @@ with the positive sign if $G$ is orientation-preserving, and the negative sign o
Proof -Let $E$ be an open domain of integration such that $\mathrm{supp}(\omega) \subseteq E \subseteq\overline{E}\subseteq V$. Since diffeomorphism take interiors to interiors, boundaries to boundaries, and sets of measure zero to sets of measure zero, $D = G^{-1} (E) \subseteq U$ is an open domain of integration containing $\mathrm{supp}(G^* \omega)$. The result follows from the previous proposition. +Let $E$ be an open domain of integration such that $\operatorname{supp}(\omega) \subseteq E \subseteq\overline{E}\subseteq V$. Since diffeomorphism take interiors to interiors, boundaries to boundaries, and sets of measure zero to sets of measure zero, $D = G^{-1} (E) \subseteq U$ is an open domain of integration containing $\operatorname{supp}(G^* \omega)$. The result follows from the previous proposition.
@@ -9557,12 +9557,12 @@ with the positive sign for a positively oriented chart, and the negative sign ot -Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\omega$ be an $n$-form on $M$. Then $\int_M \omega$ does not depend on the choice of smooth chart whose domain contains $\mathrm{supp}(\omega)$. +Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\omega$ be an $n$-form on $M$. Then $\int_M \omega$ does not depend on the choice of smooth chart whose domain contains $\operatorname{supp}(\omega)$.
Proof -Suppose $(U,\varphi)$ and $(\tilde{U},\tilde{\varphi})$ are two smooth charts such that $\mathrm{supp}(\omega)\subseteq U \cap \tilde{U}$. If both charts are positively oriented or both are negatively oriented, then $\tilde{\varphi}\circ\varphi^{-1}$ is an orientation-preserving diffeomorphism from $\varphi(U\cap\tilde{U})$ to $\tilde{\varphi}(U\cap\tilde{U})$, implying That +Suppose $(U,\varphi)$ and $(\tilde{U},\tilde{\varphi})$ are two smooth charts such that $\operatorname{supp}(\omega)\subseteq U \cap \tilde{U}$. If both charts are positively oriented or both are negatively oriented, then $\tilde{\varphi}\circ\varphi^{-1}$ is an orientation-preserving diffeomorphism from $\varphi(U\cap\tilde{U})$ to $\tilde{\varphi}(U\cap\tilde{U})$, implying That $$ \begin{align*} @@ -9578,7 +9578,7 @@ If the charts are opposiely oriented, then the two definitions of the integral o -Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\omega$ be a compactly supported $n$-form on $M$. Suppose $\Set{U_i}$ is a finite open cover of $\mathrm{supp}(\omega)$ by domains of positively or negatively oriented smooth charts, and let $\Set{\psi_i}$ be a subordinate smooth partition of unity. Define the integral of $\omega$ over $M$ to be +Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\omega$ be a compactly supported $n$-form on $M$. Suppose $\Set{U_i}$ is a finite open cover of $\operatorname{supp}(\omega)$ by domains of positively or negatively oriented smooth charts, and let $\Set{\psi_i}$ be a subordinate smooth partition of unity. Define the integral of $\omega$ over $M$ to be $$ \int_M \omega = \sum_i \int_M \psi_i \omega @@ -9601,7 +9601,7 @@ Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\o
Proof -Suppose $\Set{\tilde{U}_j}$ is another finite open cover of $\mathrm{supp}(\omega)$ by domains of positively or negatively oriented smooth charts, and $\Set{\tilde{\psi}_j}$ is a subordinate smooth partition of unity. For each $i$, we compute +Suppose $\Set{\tilde{U}_j}$ is another finite open cover of $\operatorname{supp}(\omega)$ by domains of positively or negatively oriented smooth charts, and $\Set{\tilde{\psi}_j}$ is a subordinate smooth partition of unity. For each $i$, we compute $$ \begin{align*} @@ -9651,7 +9651,7 @@ $$ Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $\omega$ be a compactly supported $n$-form on $M$. Suppose $D_1,\dots,D_k$ are open domains of integration in $\R^n$, and for $i=1,\dots,k$ we are given smooth maps $F_i : \overline{D}_i \to M$ satisfying 1. $F_i$ restricts to an orientation-preserving diffeomorphism from $D_i$ onto an open subset $W_i\subseteq M$ 2. $W_i \cap W_j = \emptyset$ when $i\neq j$ -3. $\mathrm{supp}(\omega) \subseteq \bigcup_{i=1}^k \overline{W}_i$ +3. $\operatorname{supp}(\omega) \subseteq \bigcup_{i=1}^k \overline{W}_i$ Then @@ -9745,7 +9745,7 @@ where $\partial M$ has the induced orientation such that $\iota_{\partial M}^* \ Proof **Case 1:** $M = \mathbb{H}^n$
-Because $\omega$ has compact support, there is a number $r > 0$ such that $\mathrm{supp}(\omega)$ is contained in the rectangle $A = [-r,r]^{n-1} \times [0,r]$. We can write $\omega$ in standard coordinates As +Because $\omega$ has compact support, there is a number $r > 0$ such that $\operatorname{supp}(\omega)$ is contained in the rectangle $A = [-r,r]^{n-1} \times [0,r]$. We can write $\omega$ in standard coordinates As $$ \omega = \sum_{i=1}^n \omega_i \;\mathrm{d}x^1 \wedge\cdots\wedge \hat{\mathrm{d}x}^i \wedge\cdots\wedge\mathrm{d}x^n @@ -9819,7 +9819,7 @@ $$ By the computation above, this is equal to $\int_{\partial\mathbb{H}^n} (\varphi^{-1})^* \omega$, where $\partial\mathbb{H}^n$ is given the induced orientation. Since $\mathrm{d}\varphi$ takes outward-pointing vectors on $\partial M$ to outward-pointing vectors on $\mathbb{H}^n$, it follows that $\varphi|_{U\cap\partial M}$ is an orientation-preserving diffeomorphism onto $\varphi(U)\cap\partial\mathbb{H}^n$, and this $\int_{\partial\mathbb{H}^n} (\varphi^{-1})^* \omega = \int_{\partial M} \omega$. For a negatively oriented smooth boundary chart, the same argument applies with an additional negative sign on each side of the equation. For an interior chart, we get the same computations with $\mathbb{H}^n$ replaced by $\R^n$. -Finally, let $\omega$ be an arbitrary compactly supported smooth $(n-1)$-form. Choosing a cover of $\mathrm{supp}(\omega)$ by finitely many domains of positively or negatively oriented smooth charts $\Set{(U_i, \varphi_i)}$, and choosing a subordinate smooth partition of unity $\Set{\psi_i}$, we can apply the preceding argument to $\psi_i \omega$ for each $i$ and obtain +Finally, let $\omega$ be an arbitrary compactly supported smooth $(n-1)$-form. Choosing a cover of $\operatorname{supp}(\omega)$ by finitely many domains of positively or negatively oriented smooth charts $\Set{(U_i, \varphi_i)}$, and choosing a subordinate smooth partition of unity $\Set{\psi_i}$, we can apply the preceding argument to $\psi_i \omega$ for each $i$ and obtain $$ \begin{align*} diff --git a/content/notes/math/measure_theory.mdx b/content/notes/math/measure_theory.mdx index 1b39bb6..3b2bcc7 100644 --- a/content/notes/math/measure_theory.mdx +++ b/content/notes/math/measure_theory.mdx @@ -980,14 +980,14 @@ Suppose $(X,\mathcal{T})$ is a topological space and let $\mathcal{A} = \sigma(\ Suppose $(X,\mathcal{A}, \mu)$ is a Borel measure space. The support of $\mu$ is defined as $$ - \mathrm{supp}(\mu):=\Set{x\in X | \mu(U)>0\textrm{ for every open neighbourhood }U\textrm{ of }x } + \operatorname{supp}(\mu):=\Set{x\in X | \mu(U)>0\textrm{ for every open neighbourhood }U\textrm{ of }x } $$ -The set $\mathrm{supp}(\mu)$ is closed. +The set $\operatorname{supp}(\mu)$ is closed.
Proof -Let $A=\mathrm{supp}(\mu)$. For $x\in A^c$, there exists an open neighbourhood $V_x$ of $x$ such that $\mu(V_x) = 0$. If $y\in V_x$, then $V_x$ is also an open neighbourhood of $y$, so $y\in A^c$. Hence $V_x\subseteq A^c$ for every $x\in A^c$ and so $A^c$ is open. +Let $A=\operatorname{supp}(\mu)$. For $x\in A^c$, there exists an open neighbourhood $V_x$ of $x$ such that $\mu(V_x) = 0$. If $y\in V_x$, then $V_x$ is also an open neighbourhood of $y$, so $y\in A^c$. Hence $V_x\subseteq A^c$ for every $x\in A^c$ and so $A^c$ is open.
diff --git a/content/notes/math/stochastic_analysis.mdx b/content/notes/math/stochastic_analysis.mdx index 85f44d8..af462df 100644 --- a/content/notes/math/stochastic_analysis.mdx +++ b/content/notes/math/stochastic_analysis.mdx @@ -25,7 +25,7 @@ Note that the continuity and initial value assumptions implies that $\mu$ has no Note that the decomposition $\mu = \mu_+ - \mu_-$ as the difference between two positive measures is not unique. However, it is unique when constrained to $$ - \mathrm{supp}(\mu_+) \cap\mathrm{\mu_-} = \emptyset + \operatorname{supp}(\mu_+) \cap\mathrm{\mu_-} = \emptyset $$ The uniqueness of such a composition follows the identity $\mu_+ (B) = \sup\Set{\mu(C) | C\subset B, C \text{ is a Borel set}}$. To show existence, write $\mu = \tilde{\mu}_+ - \tilde{\mu}_-$ for some positive measures $\tilde{\mu}_+$ and $\tilde{\mu}_-$. Then $\tilde{\mu}_+$ (respectively $\tilde{\mu}_-$) is absolutely continuous with respect to $\tilde{\mu} = \tilde{\mu}_+ + \tilde{\mu}_-$. By the Radon-Nikodym theorem, $\tilde{\mu}_+$ has a density $\lambda_+ (t)$ (respectively $\lambda_- (t)$) with respect to $\tilde{\mu}$. Then, the choice @@ -37,7 +37,7 @@ $$ \end{align*} $$ -gives the expected decomposition. Letting $S_+$ (respectively $S_-$) denote $\mathrm{supp}(\mu_+)$ (respectively $\mathrm{supp}(\mu_-)$) and $|\mu| = \mu_+ + \mu_-$, we get +gives the expected decomposition. Letting $S_+$ (respectively $S_-$) denote $\operatorname{supp}(\mu_+)$ (respectively $\operatorname{supp}(\mu_-)$) and $|\mu| = \mu_+ + \mu_-$, we get $$ \frac{\mathrm{d}\mu}{\mathrm{d}|\mu|} = \mathbf{1}_{S_+} - \mathbf{1}_{S_-}