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I'm trying to understand the construction of the dirac operator on a manifold, but actually I guess that doesn't really matter for the question at stake. I'm interested in understanding a definition of the principal symbol. Specifically, In Lawson and Michelsohn's Spin Geometry page 113 it says:

"Racall that the principal symbol of a differential operator $D:\Gamma (E) \to \Gamma (E)$ is a map which associates to each point $x \in X $ and each cotangent vector $\xi \in T^*_x(X)$, a linear map $\sigma _{\xi}(D):E_x \to E_x$ defined as follows. If in local coordinates we have $$ D=\sum_{|\alpha|\leq m}A_{\alpha}(x)\frac{\partial ^{|\alpha|}}{\partial x^{\alpha}} \text{ and } \xi=\sum_k \xi_k dx_k$$ where m is the order of $D$, then $$\sigma_{\xi}(D) = i^m \sum_{|\alpha|= m} A_{\alpha}(x)\xi^{\alpha}."$$ After going to the some other chapter you find out that $E$ is a complex vector bundle over $X$, a riemannian manifold, with a local trivialization $E|_U\to U \times \mathbb{C}^q$ and $A_{\alpha}(x)$ is a $q\times q$-matrix of smooth complex-valued functions.

So the question I have is if one is working with real vector bundles how does one define the principal symbol. I mean, if now one has that $A_{\alpha}(x)$ is a $q\times q$-matrix of smooth real-valued functions, how do you define the linear map $\sigma _{\xi}(D):E_x \to E_x$, because just taking the "$i^m$" factor off from the definition seems quite arbitrary. Any clarification is highly apreciated!

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Let $E\rightarrow M$ and $F\rightarrow M$ be vector bundles with spaces $\Gamma(E)$ and $\Gamma(F)$ of smooth sections. Consider a linear partial differential operator of order $k$, which is a map \begin{align} L=\sum_{|\alpha|\leq k}\ell_\alpha\partial^\alpha:\Gamma(E)&\rightarrow\Gamma(F)\\ S&\mapsto L(S)=\sum_{|\alpha|\leq k}\ell_\alpha\partial^\alpha S. \end{align} Here $\alpha=(\alpha_1,...,\alpha_m)$ is a multi-index and each $\ell_\alpha:E\rightarrow F$ is a bundle homomorphism. Now let $\omega=\omega_i\text{d}x^i\in\Gamma(T^*M)$ be a covector field (a $1$-form). The total symbol of a linear partial differential operator $L$ in the direction of the covector field $\omega$ is the bundle homomorphism: \begin{align} \sigma_L(\omega)=\sum_{|\alpha|\leq k}\omega^\alpha\ell_\alpha:E&\rightarrow F\\ e&\mapsto\sigma_L(\omega)e=\sum_{|\alpha|\leq k}\omega^\alpha\ell_\alpha e. \end{align} Here $\omega^\alpha=\omega_1^{\alpha_1}\cdots\omega_m^{\alpha_m}$. The principal symbol simply takes the highest-order partial derivative terms of the symbol, and is the bundle homomorphism: \begin{align} \hat{\sigma}_L(\omega)=\sum_{|\alpha|=k}\omega^\alpha\ell_\alpha:E&\rightarrow F\\ e&\mapsto\hat{\sigma}_L(\omega)e=\sum_{|\alpha|=k}\omega^\alpha\ell_\alpha e. \end{align} Hence, the principal symbol captures the properties of the linear partial differential operator which are held in the highest-order partial derivative terms. A linear partial differential operator is elliptic if it's principal symbol is a linear-space isomorphism for all nonzero covector fields $\omega\neq0\in\Gamma(T^*M)$.

These notions also hold for nonlinear partial differential operators between spaces of sections of vector bundles, by considering the operator's linearisation. The linearisation of a nonlinear partial differential operator is a linear partial differential operator. The symbol (principal symbol) of a nonlinear partial differential operator is the symbol (principal symbol) of its linearisation. A nonlinear partial differential operator is elliptic if the principal symbol of its linearisation is a linear space ismorphism for all nonzero covector fields $\omega\neq0\in\Gamma(T^*M)$.

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The principle symbol arise naturally when you take the Fourier transform, where the symbol appears the top order multiplier. So the choice of including $i$ is immaterial since then $i^{m}$ is a constant. The important thing is the property of $\sigma(D)$ (like whether it is elliptic, hyperbolic, invertible, etc), and that would not be changed by multiplying a constant.

I do not have Spin geometry with me, but I think what you wrote is incorrect. Here the $\sigma(D)$ should only include the top order term $|\alpha|=m$. What you wrote is the definition of the symbol instead.

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  • $\begingroup$ yeah you're right, there's a typo, the definition of $\sigma(D)$ is as you say. I'll edit it. Thanks for the answer,at the end I came to a similar conclusion! $\endgroup$ – Bruce Wayne Mar 11 '14 at 12:24

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