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Let $V$ be an even-dimensional euclidean vector space, $S$ the spinor module and $E$ a module of the Clifford algebra $C(V)$. Then $$W=\mathrm{Hom}_{C(V)}(S,E)$$ is called the twisting space of $E$. Now here is the issue:

In the book Heat Kernels and Dirac Operators the elements of $W$ are defined to super-commute with the Clifford action$^1$, but I think that they should be defined to commute with Clifford action:

The function $T\in L(S\otimes W,E)$ defined by $$T(s\otimes w)=w(s)$$ is even and hence it super-commutes with the Clifford action if and only if it commutes with the Clifford action. And this is the case if and only if all elements of $W$ commute with the Clifford action:

Fix $x\in C(V)$, $s\in S$ and $w\in W$, then $$(T\circ c_{S\otimes W}(x))(s\otimes w)=T(c_S(x)\otimes 1(s\otimes w))=T(c_S(x)(s)\otimes w)=(w\circ c_S(x))(s)$$and$$(c_E(x)\circ T)(s\otimes w)=(c_E(x)\circ w)(s)$$and hence $T\circ c_{S\otimes W}(x)=c_E(x)\circ T$ if and only if $\forall w\in W:w\circ c_S(x)=c_E(x)\circ w$.

$^1$ To be precise, the elements of $\mathrm{End}_{C(V)}(E)$ are explicitly defined to super-commute with the Clifford action and the exact definition $\mathrm{Hom}_{C(V)}(S,E)$ is not included in the book.

Filippo
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  • This is what I've also confused. I also agree with your opinion. It does NOT supercommute at all – ChoMedit Nov 27 '23 at 03:17
  • I think the way to tackle this is considering W have a trivial grading.. but I;m not sure every thing is okay – ChoMedit Nov 27 '23 at 03:18
  • @ChoMedit "I think the way to tackle this is considering W have a trivial grading" - Can you elaborate? I think that $T$ is even w.r.t. the grading discussed here. – Filippo Nov 27 '23 at 04:17
  • What I'd thought is that let the grading of elements in $W$ be one. However, I think this idea is not that good enough, since we require $W$ be a supergraded. – ChoMedit Nov 27 '23 at 07:53
  • @ChoMedit Note that proposition $3.27$ says that $E$ is isomorphic to $S\otimes W$ as a $\mathbb{Z}2$-graded complex $C(V)$-module, so $T\in\mathrm{Hom}{C(V)}(S\otimes W,E)$ should be even and we achieve that by considering the before-mentioned grading on $W$ and the tensor-product grading on $S\otimes W$. – Filippo Nov 27 '23 at 08:08
  • Yes, so we don't need any presumed(I mean, not from the definition) grading on $W$. I think your story and conclusion in the post is right. – ChoMedit Nov 27 '23 at 08:16
  • @ChoMedit Thank you for the confirmation! – Filippo Nov 27 '23 at 08:27
  • I come back because I found it very strange.. In the proof of Gauss-Bonnet-Chern by using ASindex, it uses a supercommuting elements.. $W$ should be a set of supercommuting elements. It seems really critical, I couldn't catch what's right. – ChoMedit Nov 30 '23 at 01:48
  • Perhaps, we need to choose another Clifford action on $W \otimes S$ to make everything right. – ChoMedit Nov 30 '23 at 01:52
  • @ChoMedit Can you tell me the page number? – Filippo Nov 30 '23 at 06:16
  • It's page 148. Actually there appears two grading structures, but I'm not sure what's what. – ChoMedit Nov 30 '23 at 06:43
  • @ChoMedit I don't see a contradiction: Each grading on $E$ induces a grading on $W=\mathrm{Hom}_{C(V)}(S,E)$. Since we consider two different gradings on $E$, it is not surprising that we obtain two different gradings on $W$. – Filippo Nov 30 '23 at 10:34
  • @ChoMedit If we consider the grading defined by the chirality operator, then $W$ is purely even, because the elements of $W$ commute with $C(V)$ and hence in particular with the grading (the chirality operator). – Filippo Nov 30 '23 at 10:36
  • Yeah.. the grading is not a problem. The problem seems due to the definition of $W$. – ChoMedit Nov 30 '23 at 11:00
  • How about moving this discussion to the chat? I think there is a subtle point around $W$ and archiving those points are useful for future. – ChoMedit Nov 30 '23 at 11:04
  • @ChoMedit Let us continue this discussion in chat. – Filippo Nov 30 '23 at 11:09
  • @ChoMedit May I raise your attention to a related issue? I think that in the definition of Clifford super-connections the square brackets should be interpreted as a commutators, not super-commutators. I have explained my reasoning here. I would like to hear your opinion. – Filippo Dec 01 '23 at 02:58
  • I am also very confused with this. Assume one defined $\text{Hom}{C(V)}(S, E)$ to be the space of complex linear maps $f: S\to E$ such that $f(c(v)x)=c(v)f(x)$ for any $x\in S$, then by Schur's lemma one would conclude $\text{Hom}{C(V)}(S, S)={\mathbb C}$ and in general $\text{Hom}{C(V)}(S, E)={\mathbb C}^d$, and there seems to have no clear way to define a bi-grade on $W=\text{Hom}{C(V)}(S, E)$. I have no idea if this is the way it should be. – Yuval Jan 03 '24 at 19:36
  • If take the above interpretation, on the left of (3.10) it is just the ordinary trace, on the right it is a block-diagonal matrix with each block a multiple of $\Gamma$, hence taking supertrace in this case is just like taking ordinary trace with no $\Gamma$, so (3.10) can be "proved", but this looks shaky to me (why need the "super" machinery?). – Yuval Jan 03 '24 at 20:58
  • @Yuval I think that $\mathrm{Hom}{C(V)\otimes\mathbb{C}}(S,E)$ inherits a grading from $L{\mathbb C}(S,E)$, see this answer. Does this resolve the issue or do you still see a contradiction? – Filippo Jan 04 '24 at 12:31
  • @Yuval Concerning (3.10): Let $T\in\mathrm{Hom}{C(V)\otimes\mathbb{C}}(S\otimes W,E)$ be the isomorphism from my answer. Note that the isomorphism of complex algebras$$\mathrm{End}(W)\to\mathrm{End}{C(V)\otimes\mathbb{C}}(E)$$maps $F$ to $T(1\otimes F) T^{-1}$, so (3.10) should really be read as$$2^{n/2}\mathrm{Str}_W(F)=\mathrm{Str}_E(c_E(\Gamma)T(1\otimes F) T^{-1}).$$ But this is easy since $$2^{n/2}=\mathrm{Str}_S(c_S(\Gamma))$$where $\Gamma\in C(V)\otimes \mathbb{C}$ is the chirality operator and hence – Filippo Jan 04 '24 at 13:11
  • $$2^{n/2}\mathrm{Str}W(F)=\mathrm{Str}{S\otimes W}(c_S(\Gamma)\otimes F)=\mathrm{Str}_E(T(c_S(\Gamma)\otimes 1)(1\otimes F)T^{-1})=\mathrm{Str}_E(c_E(\Gamma)T(1\otimes F)T^{-1})$$ – Filippo Jan 04 '24 at 13:11

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