The Stacks project

Remark 15.70.6. In the yoga of super vector spaces the sign used in the proof of Lemma 15.70.5 above can be explained as follows. A super vector space is just a finite dimensional vector space $V$ which comes with a direct sum decomposition $V = V^+ \oplus V^-$. Here we think of the elements of $V^+$ as the even elements and the elements of $V^-$ as the odd ones. Given two super vector spaces $V$ and $W$ we set

\[ (V \otimes W)^+ = (V^+ \otimes W^+) \oplus (V^- \otimes W^-) \]

and similarly for the odd part. In the category of super vector spaces the isomorphism

\[ \psi : V \otimes W \longrightarrow W \otimes V \]

is defined to be the usual one, except that on the summand $V^- \otimes W^-$ we use the negative of the usual identification. In this way we obtain a symmetric monoidal category, see Categories, Section 4.42. An object $V$ of the category of super vector spaces has a left dual which we denote $V^\vee $ which comes equipped with an identity $\eta : \mathbf{1} \to V \otimes V^\vee $ and an evaluation map $\epsilon : V^\vee \otimes V \to \mathbf{1}$ which induce canonical isomorphisms $\mathop{\mathrm{Hom}}\nolimits (V, W) = W \otimes V^\vee $ and $\mathop{\mathrm{Hom}}\nolimits (V^\vee , U) = V \otimes U$, see Categories, Lemma 4.42.6. Given three super vector spaces $U$, $V$, $W$ we can try to construct the analogue

\[ c : \mathop{\mathrm{Hom}}\nolimits (V, W) \otimes U \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (U, V), W) \]

of the maps $c_{p, r, s}$ which occur in the lemma above. Using the formulae given above (which do not involve signs) this becomes a map

\[ W \otimes V^\vee \otimes U \longrightarrow W \otimes (V \otimes U^\vee )^\vee = W \otimes (U^\vee )^\vee \otimes V^\vee \]

To find this arrow in a canonical fashion we need to do two things:

  1. we need to use the commutativity constraint $\psi : V^\vee \otimes U \to U \otimes V^\vee $ which introduces a sign on $(V^\vee )^- \otimes U^-$, and

  2. we need to use the canonical isomorphism $U \to (U^\vee )^\vee $ which comes from the identification of $U^\vee $ as the right dual of $U$ using $\psi $ as in Categories, Lemma 4.42.10. This differs from the usual identification by $-1$ on the odd part of $U$.

Part (1) explains the sign $(-1)^{qr}$ in the proof of the lemma and part (2) explains the sign $(-1)^ r$ in the proof of the lemma.

Comments (2)

Comment #4343 by Manuel Hoff on

In the sentence "The category of super vector spaces has an internal hom which we denote .", i think "internal hom" should be replaced by something like "duals".

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A61. Beware of the difference between the letter 'O' and the digit '0'.