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Let $M$ be a smooth manifold and $\mathfrak{X}(M)$ the real vector space consisting of vector fields over $M$. Let $TM\boxtimes TM$ be the vector bundle over $M\times M$ whose fiber at $(x,y)\in M\times M$ is $T_xM\otimes T_yM$. Finally, let $\Gamma(TM\boxtimes TM)$ denote the sections of this vector bundle.

Is there any essential difference between the tensor product of real vector spaces $\mathfrak{X}(M)\otimes_{\mathbb{R}}\mathfrak{X}(M)$ and the real vector space $\Gamma(TM\boxtimes TM)$?

There seems to be a well-defined $\mathbb{R}$-linear map $\phi:\mathfrak{X}(M)\otimes_{\mathbb{R}}\mathfrak{X}(M)\rightarrow \Gamma(TM\boxtimes TM)$ given by $\phi(\sum_k X_k\otimes Y_k)(x,y)=\sum_k X_k(x)\otimes Y_k(y)$. Provided that the bundle $TM\boxtimes TM$ is paracompact and can be covered by finitely many bundle charts, I think that you can use a partition of unity to show that $\phi$ is surjective. I'm having a bit of trouble proving it's injective though. I also can't think of a non-trivial element of $\mathfrak{X}(M)\otimes_{\mathbb{R}}\mathfrak{X}(M)$ that would map to the zero section.

Josh Burby
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    In $\mathfrak{X}(M)\otimes\mathfrak{X}(M)$ you should make explicit what ring is used when computing the tensor product. – Mariano Suárez-Álvarez Apr 03 '13 at 03:17
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    The tensor product of vector bundles is the tensor product of $C^{\infty}(M)$-modules, and it's not the tensor product of underlying vector spaces. – Qiaochu Yuan Apr 03 '13 at 03:18
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    @QiaochuYuan Could you elaborate a bit on your comment? Are you saying $\Gamma(TM\otimes TM)=\mathfrak{X}(M)\otimes_{C^\infty(M)}\mathfrak{X}(M)$ and $\Gamma(TM\otimes TM)\neq\mathfrak{X}(M)\otimes_{\mathbb{R}}\mathfrak{X}(M)$? – Josh Burby Apr 03 '13 at 04:34
  • @Josh: yes, that's what I'm saying. – Qiaochu Yuan Apr 03 '13 at 05:54
  • @Qiaochu: could you please provide a proof or reference for your statement about the tensor product over $C^\infty(M) $ ? – Georges Elencwajg Apr 03 '13 at 08:05
  • @Georges: hmm. Well, if you don't think it's obvious then I'm probably wrong, but doesn't this follow from the Serre-Swan theorem together with the observation that the two tensor products preserve colimits and do the same thing to free modules resp. trivial bundles? (I guess I need $M$ compact.) – Qiaochu Yuan Apr 03 '13 at 08:17
  • Dear @Qiaochu, I think your result is correct and that indeed it could follow from some non-compact version of Serre-Swan. However I thought there might be some slick proof involving the vanishing of cohomology for the sheaf of sections of $C^\infty$-vector bundles that would prove your assertion and simultaneously the homologous ones ones for affine algebraic varieties or Stein manifolds, where cohomology of vector bundles also vanishes. So my comment was prompted by the hope that you had something like that in mind, and not at all because I doubted the truth of your result. – Georges Elencwajg Apr 03 '13 at 08:53
  • @QiaochuYuan: I'm not familiar with the Serre-Swan theorem, but isn't $\Gamma(TM\otimes TM)$ a module over $C^\infty(M\times M)$ while $\mathfrak{X}(M)\otimes_{C^\infty(M)}\mathfrak{X}(M)$ is a module over $C^\infty(M)$? I think I might be using a non-standard notation for what I'm calling $TM\otimes TM$ - the base is $M\times M$ rather than $M$. – Josh Burby Apr 03 '13 at 20:16
  • @Josh: oh, sorry, I misread the definition of that bundle. – Qiaochu Yuan Apr 03 '13 at 20:17
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    @JoshBurby $TM\otimes TM$ denotes, in most contexts, the bundle on $M$ whose fiber at each point $p$ is $T_pM\otimes T_pM$. The tensor product which lives on $M\times M$ is then denoted $TM\boxtimes TM$ or something like that (and it is $p_1^(TM)\otimes p_2^(TM)$, with the $p_i$ the projections $M\times M\to M$) – Mariano Suárez-Álvarez Apr 03 '13 at 21:02
  • @MarianoSuárez-Alvarez Thanks very much for clearing that up - I changed the notation in the post. – Josh Burby Apr 03 '13 at 21:32

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Let $M=\def\RR{\mathbb R}\RR$. Let $\def\X{\mathfrak X}\X(M)$ be the vector space of tangent fields to $M$, and let $\partial\in\X(M)$ be any vector field which is never-zero, so that $\{\partial\}$ is a $C^\infty(M)$-basis of the $C^\infty(M)$-module $\X(M)$, and the map $$\alpha:f\in C^\infty(M)\mapsto f\partial\in\X(M)$$ is an isomorphism.

The bundle $TM$ is a trivial line bundle on $M$. It follows that $TM\boxtimes TM$ is also a trivial line bundle on $M\times M$. We can view $\partial$ as a section $M\to TM$ which generates at each point $TM$, and then $(p,q)\in M\times M\mapsto\partial_p\otimes\partial_q\in (TM\boxtimes TM)_{(p,q)}$ is a section of $TM\boxtimes TM$ which generates at each point, let us call it $D$. It follows that $$\beta:f\in C^{\infty}(M\times M)\mapsto fD\in\Gamma(TM\boxtimes TM)$$ is a vector space isomorphism.

Now your map $\X(M)\otimes_\RR\X(M)\to\Gamma(TM\otimes TM)$ can be conjugated by the maps $\alpha$ and $\beta$ to give a map $$C^\infty(M)\otimes_\RR C^\infty(M)\to C^\infty(M\times M).$$

Make it explicit and see that it is not an isomorphism.

The problem of deciding when a function is in the image of the above map is a very classical one. Here is a necessary condition: suppose $h(x,y)=\sum_{\ell=1}^nf_i(x)g_i(y)$ can be written as a sum of $n$ products of decomposable functions. Then the determinant $$\mathcal W_n(h)=\det\left(\frac{\partial^{i+j}h}{\partial x^i\partial y^j}\right)_{i,j=0,\dots,n}$$ vanishes. This determinant is called a Wronksian, if I recall correctly (although it is not the same as the Wronskian one finds in the context of linear ODEs —it is related, of course)

Now, a little work will show that $\mathcal W_n(\exp(xy))\neq0$ for all $n\geq1$, so that $\exp(xy)$ is not in the image of our map. (In fact, $\mathcal W_n(\exp(xy))$ is $\exp((n+1)xy)$ times the Vandermonde determinant for $1$, $2$, $\dots$, $n+1$.)

Mariano Suárez-Álvarez
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  • (Notice the problem is with surjectivity) – Mariano Suárez-Álvarez Apr 03 '13 at 21:36
  • The image seems to be finite sums of functions of the form $h(x,y)=f(x)g(y)$, which is a vector subspace of $C^\infty(M\times M)$. I suspect something like $\sin(xy)$ does not live in this space, but I have to admit I am having a hard time proving it. – Josh Burby Apr 03 '13 at 22:35
  • This is great - thank you! So the "obvious map" is not an isomorphism, although it feels like it is nearly so. I wonder if, when $M$ is compact at least, $\phi$ can be extended to an isomorphism between the completions of $\mathfrak{X}(M)\otimes_{\mathbb{R}}\mathfrak{X}(M)$ and $\Gamma(TM\boxtimes TM)$ under some norm. Maybe something like the Stone-Weierstrass approximation theorem would do it. – Josh Burby Apr 04 '13 at 01:20
  • If you complete (with respect to the natural topology; these are nuclear spaces) you do get isomorphisms. – Mariano Suárez-Álvarez Apr 04 '13 at 01:23