I was looking at this definition of the symbol of a differential operator, and am unsure what "$T^*X\otimes_XE$" means. I couldn’t find an explanation anywhere on nlab either. My main question I guess is what $_XE$ means. I am especially trying to reconcile this definition with another one I have seen, namely that $$\sigma(D):(T^\ast X)^{\otimes k}\to E^\ast\otimes E$$where $k$ is the dimension of $X$
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1$\begingroup$ I think it's just a notation for the usual tensor product of two vector bundles (it even says so in the nlab link, you can chase the second link to the definition). I guess $\otimes_X$ instead of just $\otimes$ is to remind that the base manifold is $X$ (and that it's a tensor product of vector bundles rather than just vector spaces over $\mathbb{R}$). $\endgroup$– M.G.Commented May 29 at 14:03
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$\begingroup$ But then that wouldn't coincide with the second definition I've written, right? Do they define different operations? $\endgroup$– Barsa JahanpanahCommented May 29 at 16:07
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1$\begingroup$ For your second definition, $D:C^{\infty}X\rightarrow C^{\infty}X$ is a differential operator of order $k$, which should be symmetric on the indices in local coordinates and its coefficients of order $k$ define its symbol as a map $\sigma(D): S^k(T^*X)\otimes E\rightarrow E$, where $S^k$ is the $k$-th symmetric power of the cotangent bundle. This is essentially the definition on the page you link to. $\endgroup$– F ZaldivarCommented May 29 at 17:30
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