6
$\begingroup$

Let's say we have an oriented compact 4-d Riemannian spin manifold $(M,g)$. Everybody who's anybody has heard about the index of the Dirac operator $D: S^+\rightarrow S-$; it's the $\hat{A}$-genus, which is $\displaystyle\frac{-1}{8}\tau(M)$ ($\tau$ is the signature). I don't know too much about the index theorem and it's inner workings, but I'm wondering what happens if you perturb the Dirac operator and consider $$D_{f,s}=D+s\textrm{grad}(f)\cdot$$ where s is a (let's say small) real parameter and $f\in C^{\infty}(M)$; "$\cdot$" indicates Clifford multiplication, and maps $S^+\rightarrow S^-$ since elements of $TM$ anticommute with the volume element in dimension 4.

Does the index stay the same? Change in a predictable way related to $f$? What if $f$ is special somehow? What about in the special case where $\tau=0$? Or the even-more-special case where $\tau=0$ because the dimension of the harmonic spinors is $0$, e.g. when $R>0$? How does the dimension of the kernel jump as $s$ changes?

Sorry for the avalanche of questions. I'm interested in any information people have about any subset of them.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ I'm assuming $M$ is compact, and I'm assuming $\nabla f$ makes sense as a map $S^+\to S^-$ (I don't immediately see why it is). The point is that $D$ is Fredholm, and the space of Fredholm operators is open in the (Banach) space of bounded linear operators. So $ind(D)=ind(D+K)$ for any compact operator $K$ (hence $s$ can be anything), and if $K$ is not compact then under mild assumptions $ind(D)=ind(D+sK)$ for $s$ sufficiently small. $\endgroup$
    – Chris Gerig
    Commented Apr 8, 2016 at 16:46
  • $\begingroup$ Sorry, I meant to add the compactness assumption and then I forgot. And Clifford multiplication by elements of $TM$ maps $S^+\rightarrow S^-$ since the basis vectors $e_i$ anticommute with the volume element $e_1e_2e_3e_4$ (Right? Or am I missing something? Gonna feel real dumb if I am...) $\endgroup$
    – Brian Klatt
    Commented Apr 8, 2016 at 17:12
  • $\begingroup$ You are right if you say that sgrad(f) acts by Clifford multiplication, and then @ChrisGerig's comment answer your question. I suggest you add that small detail to your question. $\endgroup$
    – Sebastian Goette
    Commented Apr 8, 2016 at 17:27
  • $\begingroup$ Okay, my apologies that it was unclear; the "$\cdot$" after $\textrm{grad}(f)$ was intended to indicate Clifford multiplication. $\endgroup$
    – Brian Klatt
    Commented Apr 8, 2016 at 17:37
  • $\begingroup$ @ChrisGerig Is it clear that Clifford multiplication by $s \textrm{grad}(f)$ is a compact operator? $\endgroup$
    – Brian Klatt
    Commented Apr 8, 2016 at 17:41

2 Answers 2

Reset to default
5
$\begingroup$

For this particular case the two operators are conjugate (though note the conjugacy is not unitary unless $s$ is imaginary)

$$ D_{f,s}=e^{-sf}D e^{sf} $$ so that the dimension of both the kernel and cokernel are independent of $s$. Note that Chris Gerig's comments on the other hand apply more generally. The index of $$ D_{s,\theta}=D+s\theta⋅ $$ for any one-form $\theta$ is independent of $s$.

Improve this answer
$\endgroup$
4
$\begingroup$

Making my comment a formal answer: The perturbation object is a compact operator (for any scaling $s$), and $D$ is Fredholm. The space of Fredholm operators is open in the (Banach) space of bounded linear operators, and moreover $ind(D)=ind(D+K)$ for any compact operator $K$.
The question about spectral flow (jumps in the kernel) is much more intricate, it should depend at least on the critical points of $f$ and the magnitude of $s$. But again by general facts of Functional Analysis, for $|s|$ sufficiently small the dimension of the kernel won't jump (and in general it will only jump for a discrete set of $s\in\mathbb{R}$).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ It might help to specify the setup in which these operators are Fredholm, see Brian's questions above. $\endgroup$
    – Sebastian Goette
    Commented Apr 26, 2016 at 9:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .