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

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

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

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

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

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

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

Comment #4493 by Johan on