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What I want to ask is the multiplicativity of the analytic index of a family of Dirac operators.

In the single operator case the analytic index of elliptic operator is multiplicative. This is proved in an Atiyah-Singer paper (ASI) and also in some other books like Lawson-Michelsohn's Spin geometry, where the kernels of the elliptic operator are not assumed to form a bundle. Of course the analytic index for family of elliptic operators is (and should be) also multiplicative, but I kind of ask for more.

For the family case, let's say we have two fibrations $\pi_1:M_1\to B_1$ and $\pi_2:M_2\to B_2$ with closed spin fibers of even dimension. Form the product $\pi:M_1\times M_2\to B_1\times B_2$, and the metrics and connections needed to define the spin Dirac operator are in product form. Define the spin Dirac operator $D$ for $\pi$ by $$D=D_1\otimes\textrm{id}\oplus\textrm{id}\otimes D_2,$$ where $D_k$ is the spin Dirac operator for $\pi_k:M_k\to B_k$ and the tensor product above is the external tensor product (of course we can twist the spinor bundle of $\pi_k$ by a complex vector bundle, but I don't bother to do this here). Let's assume the kernels of $D_1$, $D_2$ and $D$ do form a bundle. The multiplicativity of the analytic index in the family case leads me to think that we should have something like $$\ker(D)\cong\ker(D_1)\otimes\ker(D_2).$$ Again the above tensor product is the external tensor product (and $\mathbb{Z}_2$-graded).

But when I google the kernel of tensor product of two linear maps (on tensor product of finite dimensional vector spaces, say), mathstackexchange gives a similar formula but the right-hand side is a little bit different.

So is my guess correct or not? If it's correct is there a proof (I google and can't find any), and in particular, I want to know whether the isomorphism can be written explicitly or not. If it's wrong what is the relation of $\ker(D)$ with $\ker(D_1)$ and $\ker(D_2)$?

Thanks.

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  • $\begingroup$ Could you provide a link to the mathstackexchange question, please? $\endgroup$ – Sebastian Goette Jul 13 '16 at 7:30
  • $\begingroup$ There are some. In particular, this one math.stackexchange.com/questions/541541/… $\endgroup$ – Ho Man Ho Jul 13 '16 at 16:00
  • $\begingroup$ I hope I have not misunderstood the meaning there $\endgroup$ – Ho Man Ho Jul 13 '16 at 16:01
  • $\begingroup$ The problem is your first display, where $D_1$ and $D_2$ seem to commute. Then the math.stackexchange answer is correct. To get an elliptic operator, you want $D_1$ and $D_2$ to anticommute, which gives your formula for the kernel. $\endgroup$ – Sebastian Goette Aug 2 '16 at 11:20

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