2
$\begingroup$

I am struggeling with the following comment that I read regarding the de Rham complex:

Define $(d + \delta)_e : C^\infty(\Lambda^e(T^*M)) + C^\infty(\Lambda^o(T^*M))$ where

\begin{equation} \Lambda^e(T^*M) = \oplus_{k} \, \Lambda^{2k}(T^*M), \text{ and } \Lambda^o(T^*M) = \oplus_{k} \, \Lambda^{2k+1}(T^*M) \end{equation} denote the differential forms of even and odd degrees.

Here comes the statement that I don't understand:

$(d + \delta)$ is an elliptic operator since the associated Laplacian $\triangle = (d + \delta)_e^*(d + \delta)_e$ is elliptic since dim$(\Lambda^e)$ = dim$(\Lambda^o)$.

Why can I deduce the ellipticity of $\triangle$ from the fact that the dimensions of these two spaces agree ? Many thanks for any hints !

  • 0
    Anyway, see http://en.wikipedia.org/wiki/Laplacian_operators_in_differential_geometry and http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator I have only a vague memory of an argument giving ellipticity from a surprisingly meager statement about nonvanishing of something linear, but it has been a long time.2012-04-29

1 Answers 1

3

It is indeed true that the operator $d+\delta$ is elliptic. But as was already pointed out in the comments, this does not follow from the fact that the bundles on which it acts have the same dimension (consider for example the zero operator).

But it follows from the following: Denote by $\Delta$ the laplace operator, i.e., $\Delta = (d+\delta)^2 = d\delta+\delta d$.

Now you can compute that the principal symbol of $\Delta$ can be described in terms of interior and exterior multiplication of differential forms:

$\sigma_{d+\delta}(x,\xi) = int(\xi) + ext(\xi)$ and hence $\sigma_{\Delta}(x,\xi) = \sigma_{d+\delta}(x,\xi)^2 = int(\xi)ext(\xi) + ext(\xi)int(\xi)$ which can be computed to be multiplication by $\|\xi\|^2$ which is invertible outside the zero section of the cotangent bundle.

Hence both $\Delta$ as well as $d+\delta$ are elliptic.