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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?2012-04-28
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    @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 ?2012-04-28
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    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?2012-04-28
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    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).2012-04-29
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    @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 far2012-04-29
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    @WillJagy the book I am currently studying is written by Peter Gilkey, the title is "Invariance Theory, The Heat Equation And The Atiyah Singer Index Theorem"2012-04-29
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    I see one thing that's missing, your first line does not actually give any definition of $(d + \delta)_e$ or have any $\rightarrow$ sign.2012-04-29
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    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

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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.