7
$\begingroup$

I'm confused about a seemingly basic point about a classical result on positive scalar curvature and would appreciate it if an expert could help me out.

Let $M^n$ be a closed spin manifold with spinors $S$. Let $A$ be a $C^*$-algebra and $V$ be a flat Hilbert $A$-module bundle over $M$. Then one can form a twisted Dirac operator $D$ acting on sections of $S\otimes V$. This operator defines a class $[D]\in KK_n(C(M),A)$, where $C(M)$ denotes the continuous functions on $M$, and it has an index in $K_n(A)$.

(For example, this is explained in section 3 of Jonathan Rosenberg's 1986 paper C$^*$-Algebras, Positive Scalar Curvature and the Novikov Conjecture III; the real case is studied there but it seems to apply equally well to the complex case.)

It's sometimes mentioned, or at least implied, e.g. in the above paper and on p. 264 of

https://www.intlpress.com/site/pub/files/_fulltext/journals/sdg/2006/0011/0001/SDG-2006-0011-0001-a009.pdf,

that when $M$ admits a positive scalar curvature metric, the index of $D$ vanishes in $K_n(A)$. The subscript $n$ is what I'm trying to understand.

The argument (p. 332 of the first paper above) seems to be to reduce the computation to the case when $n$ is even, where the result holds by the Lichnerowicz formula (this case I understand). Now suppose $M$ is odd-dimensional. Consider the Dirac operator $D_{M\times S^1}$ on $M\times S^1$, where $S^1$ is a flat circle. Then $[D_{M\times S^1}]\in KK(C(M\times S^1),A).$

Using the embedding of the open set $M\times\mathbb{R}\hookrightarrow M\times S^1$, $[D_{M\times S^1}]$ maps to a class in $KK(C_0(M\times\mathbb{R}),A)$, which after applying the isomorphism to $KK_1(C(M),A)$ gives $[D]$. Since $M\times S^1$ also has a metric of positive scalar curvature, obtained by taking the product of the metrics on $M$ and $S^1$, the index of $[D_{M\times S^1}]$ vanishes in $K_0(A)$.

Question: How does this imply that the index of the Dirac operator $D$ on $M$ vanishes in $K_1(A)$?

It seems to me that we are taking two indices, the index in $K_0(A)$ on the larger manifold $M\times S^1$, and a more "refined" index in $K_1(A)$ on $M$, and that one may want to use the isomorphism $K_1(A)\cong K_0(A\otimes C_0(\mathbb{R}))$ somewhere.

$\endgroup$
  • $\begingroup$ I am not an expert in these things but it seems to me that the proof in "Spin geometry" by Lawson and Michelsohn is not by reduction to the even-dimensional case. Did you look there, or did you have a special reason to follow the above argument? $\endgroup$ – Igor Belegradek May 4 '18 at 18:58
  • $\begingroup$ I wasn’t aware that there was a proof involving a K-theoretic index in Spin Geometry. Do you know which chapter it is in? $\endgroup$ – ougoah May 7 '18 at 21:49
  • $\begingroup$ The proof combines several ingredient spread around the book. It does discuss KO-valued index theory. Just look in the index for "positive scalar curvature" and trace back from there. After the index is defined one applies the Lichnerowicz formula. $\endgroup$ – Igor Belegradek May 7 '18 at 22:04
  • $\begingroup$ I think you’re right. There is another way to prove this, using what you said above, which looks like it would generalise to the Hilbert A-module scenario. Still, I wonder how the $K_1$-valued index relates to the index on $M\times S^1$. $\endgroup$ – ougoah May 8 '18 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.