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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 !

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Oh, dear. Where did you read this? –  Will Jagy Apr 28 '12 at 23:18
    
@WillJagy I am sorry maybe I copied it wrongly - I don't want to put the author in a bad light, may I ask you to point out where the blundeer lies so that I can check with the original text ? –  harlekin Apr 28 '12 at 23:24
    
Actually, no, the Laplacian is elliptic. See F. Warner, Foundations of differentiable manifolds and Lie groups, or various more recent books. I just always want the context before I spend time on these questions, and that includes the book title. In particular, your language suggests something different from the Laplace-Beltrami operator. What do you think the asterisk means in their definition of $\Delta,$ and what do you think $\delta$ means? –  Will Jagy Apr 28 '12 at 23:51
    
In general just because the dimensions agree does not mean that the operator is elliptic. I don't know how to determine ellipticity without computing the symbol (which is not that bad if you know about the language of Dirac operators). –  Eric O. Korman Apr 29 '12 at 0:04
    
@WillJagy from what I understand so far, the asterisk means taking the adjoint with respect to an inner product, and in the same vein, $\delta$ stands for the adjoint to $d$ so that $(d\omega, \eta)$ = $(\omega, \delta \eta)$ for each p-form $\omega$ and $(p + 1)$-form $\eta$. Hope this is correct - sorry I haven't touched upon a lot of this theory so far –  harlekin Apr 29 '12 at 0:10

1 Answer 1

up vote 3 down vote accepted

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.

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