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.

Comment #4343 by Manuel Hoff on

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

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