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Fix a sign error in dga.tex

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aisejohan committed Apr 1, 2014
1 parent aa28d0a commit 0a09505a7ea044bc6592d4ae12b6ce12fcc0a204
Showing with 61 additions and 17 deletions.
  2. +60 −17 dga.tex
@@ -84,6 +84,7 @@ Daniel Miller
Yusuf Mustopa
Josh Nichols-Barrer
Thomas Nyberg
Masahiro Ohno
Catherine O'Neil
Martin Olsson
Brian Osserman
77 dga.tex
@@ -1367,35 +1367,78 @@ \section{Injective modules over algebras}

Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a
(right) differential graded $A$-module, then $M^\vee$ (as defined above
for graded modules) is a differential graded module as well with
differential given by the contragredient of $\text{d}_M$ up to sign.
The sign rule (see Example \ref{example-dgm-dg-cat}) is to set
{\bf left} differential graded $A$-module, then we will endow $M^\vee$
(with its graded module structure as above) with a right differential
graded module structure by setting
\text{d}_{M^\vee}(f) = - (-1)^n f \circ \text{d}_M^{-n - 1}
\quad\text{in }(M^\vee)^{n + 1}
where $f \in (M^\vee)^n = \Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$
and $\text{d}_M^{-n - 1} : M^{-n - 1} \to M^{-n}$ is the differential
of $M$. This does indeed work, namely, if $a \in A^m$, $x \in M^{-n - m - 1}$
and $f$ as above, then
for $f \in (M^\vee)^n = \Hom_\mathbf{Z}(M^{-n}, \mathbf{Q}/\mathbf{Z})$
and $\text{d}_M^{-n - 1} : M^{-n - 1} \to M^{-n}$ the differential
of $M$\footnote{The sign rule is analogous to the one in
Example \ref{example-dgm-dg-cat}, although there we are working with
right modules and the same sign rule taken there does not work for
left modules. Sigh!}.
We will show by a computation that this works.
Namely, if $a \in A^m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee)^n$,
then we have
\text{d}_{M^\vee}(f a)(x) & =
- (-1)^{n + m} (f a)(\text{d}_M(x)) \\
& =
- (-1)^{n + m} f(a\text{d}_M(x)) \\
& =
-(-1)^n f(\text{d}_M(ax) - \text{d}(a)x) \\
& =
-(-1)^n[-(-1)^n \text{d}_{M^\vee}(f)(ax) - (f\text{d}(a))(x)] \\
& =
(\text{d}_{M^\vee}(f)a)(x) + (-1)^n (f\text{d}(a))(x)
the third equality because
$\text{d}_M(ax) = \text{d}(a)x + (-1)^m a\text{d}_M(x)$.
In other words we have
$\text{d}_{M^\vee}(fa) = \text{d}_{M^\vee}(f)a + (-1)^n f \text{d}(a)$
as desired.

If $M$ is a {\bf right} differential graded module, then the sign rule above
does not work. The problem seems to be that in defining the left $A$-module
structure on $M^\vee$ our conventions for graded modules above defines $af$
to be the element of $(M^\vee)^{n + m}$ such that $(af)(x) = f(xa)$ for
$f \in (M^\vee)^n$, $a \in A^m$ and $x \in M^{-n - m}$ which in some sense
is the ``wrong'' thing to do if $m$ is odd. Anyway, instead of changing
the sign rule for the module structure, we fix the problem by using
\text{d}_{M^\vee}(f) = (-1)^n f \circ \text{d}_M^{-n - 1}
when $M$ is a right differential graded $A$-module. The computation for
$a \in A^m$, $x \in M^{-n - m - 1}$ and $f \in (M^\vee)^n$ then becomes
\text{d}_{M^\vee}(a f)(x) & =
- (-1)^{n + m} (a f)(\text{d}(x)) \\
(-1)^{n + m} (f a)(\text{d}_M(x)) \\
& =
(-1)^{n + m} f(\text{d}_M(x)a) \\
& =
- (-1)^{n + m} f(\text{d}(x)a) \\
(-1)^{n + m} f(\text{d}_M(ax) - (-1)^{m + n + 1} x\text{d}(a)) \\
& =
- (-1)^{n + m} f(\text{d}(xa) - (-1)^{n + m}x\text{d}(a)) \\
(-1)^m \text{d}_{M^\vee}(f)(ax) + f(x\text{d}(a)) \\
& =
(\text{d}(a) f)(x) + (-1)^m (a \text{d}_{M^\vee}(f))(x)
(-1)^m (a\text{d}_{M^\vee}(f))(x) + (\text{d}(a)f)(x)
in other words we have $\text{d}(af) = \text{d}(a)f + (-1)^m a \text{d}(f)$
the third equality because
$\text{d}_M(xa) = \text{d}_M(x)a + (-1)^{n + m + 1} x\text{d}(a)$.
In other words, we have
$\text{d}_{M^\vee}(af) = \text{d}(a) f + (-1)^ma\text{d}_{M^\vee}(f)$
as desired.

Similarly, if $M$ is a left differential graded module, then
$M^\vee$ becomes a right differential graded module with the same
sign rule for $\text{d}_{M^\vee}$. Computation omitted.
We leave it to the reader to show that with the conventions above
there is a natural evaluation map $M \to (M^\vee)^\vee$ in the
category of differential graded modules if $M$ is either a differential
graded left module or a differential graded right module. This works
because the sign choices above cancel out and the differentials of
$((M^\vee)^\vee$ are the natural maps $((M^n)^\vee)^\vee \to ((M^{n + 1})^\vee)^\vee$.

@@ -1649,7 +1692,7 @@ \section{I-resolutions}
\item $I = \lim I/F_pI$,
\item the inclusions $F_iI \to F_{i - 1}I$ are admissible
\item the quotients $F_{i - 1}I/F_iI$ are direct sums of the\
\item the quotients $F_{i - 1}I/F_iI$ are direct sums of the
modules $A^\vee[k]$.
In fact, condition (2) is a consequence of condition (3), see

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