Let $M$ be a smooth manifold. I have seen a Riemannian metric be defined in many ways:

  • A smooth choice of an inner product $g_p:T_pM\times T_pM\to\mathbb{R}$ which is symmetric and positive-definite, at each point $p\in M$.

  • An element of $\Gamma(T^\ast M\otimes T^\ast M)$ (which is positive-definite, symmetric)

  • An element of $\Gamma(\mathrm{Hom}(TM\otimes TM,\mathbb{R}))$ (positive-definite, symmetric)

Are these all equivalent? The reason why I wonder is that if $TM$ is not parallelisable, then $\Gamma(TM)$ is not a free module, so we don't have the isomorphism $\Gamma(T^\ast M)\otimes\Gamma(T^\ast M))\cong\mathrm{Hom}(\Gamma(TM)\otimes\Gamma(TM),C^\infty(M))$.


Can we equivalently define a Riemannian metric as an element of $\Gamma(T^\ast M)\otimes\Gamma(T^\ast M)$, or is this just utterly wrong?

  • $\begingroup$ Judging by your last sentence, is the $\Gamma$ in bullet point $3$ a typo? $\endgroup$ – snulty Jun 29 '16 at 16:25
  • $\begingroup$ It's not. Is there anything wrong with it? $\endgroup$ – B. Pasternak Jun 29 '16 at 16:30
  • $\begingroup$ I'm just trying to unpack the notation. Normally I'd write $\Gamma(E)$ is the space of sections of the bundle $\pi: E\to M$, and the M is implied. Or explicitly write $\Gamma(M,E)$. So bullet point $2$ makes sense to me. I'm still a bit new to thinking of bundles properly, so I can't tell what $\Gamma(\mathrm{Hom}(TM\otimes TM),\Bbb R)$ is. Firstly because I can't tell what $\mathrm{Hom}(TM\otimes TM)$ is: are these bundle morphisms from the tensor product bundle to itself? And should I them think of $\mathrm{Hom}(TM\otimes TM)$ as a bundle over $\Bbb R$? $\endgroup$ – snulty Jun 29 '16 at 16:47
  • $\begingroup$ There is a typo. Everything is bundles over $M$, but OP should have typed $\Gamma(\text{Hom}(TM\otimes TM,\Bbb R))$. $\endgroup$ – Ted Shifrin Jun 29 '16 at 16:53
  • $\begingroup$ @TedShifrin You were right about the typo of course, I missed that because snulty asked about the $\Gamma$. $\endgroup$ – B. Pasternak Jun 29 '16 at 19:20

A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$).

Notice that we're never trying to say that $\Gamma(E\otimes F) \cong \Gamma(E)\otimes\Gamma(F)$.

  • $\begingroup$ I agree that $\mathrm{Hom}(\Gamma(TM)\otimes\Gamma(TM),C^\infty(M))\cong(\Gamma(TM)\otimes \Gamma(TM))^\ast$, but if this is supposed to be the same as $\Gamma(T^\ast M\otimes T^\ast M)$, then isn't that wrong? $\endgroup$ – B. Pasternak Jun 29 '16 at 8:18
  • 1
    $\begingroup$ I don't know where you keep getting $\Gamma(TM)\otimes\Gamma(TM)$, when it should be $\Gamma(TM\otimes TM)$. On the other hand, it seems to me that a partition of unity argument shows that in the smooth category we can write any (global) $2$-tensor as a sum of tensor products of (global) $1$-forms. This is definitely false in the holomorphic category, for example. $\endgroup$ – Ted Shifrin Jun 29 '16 at 16:50
  • $\begingroup$ @TedShifrin Thanks for your response on the question, I wasn't trying to annoy the op but merely understand myself. On the idea $\mathrm{Hom}(E,\Bbb R)=E^*$, I've seen in spivak the construction of the dual of a bundle, which I believe this is what this is. He mentions that $E^*$ operates on $E$ in that sense that given $s\in \Gamma(E)$ and $\sigma\in \Gamma(E^*)$, that we get a map $\sigma(s):M\to \Bbb{R}$, by $\sigma(s)(p):=\sigma(p)(s(p))$. Is there a way that $E^*$ might operate on $E$ with the use of sections? $\endgroup$ – snulty Jun 29 '16 at 17:20
  • $\begingroup$ @snulty: I never thought you were trying to annoy. But what you just wrote down is precisely the way sections of $E^*$ "eat" or act on sections of $E$ and give you scalar functions. Of course, it comes a pointwise action of the dual vector space $(E_p)^*$ on the vector space $E_p$. Was there something I missed in your question? $\endgroup$ – Ted Shifrin Jun 29 '16 at 17:44
  • $\begingroup$ @TedShifrin oh I think I meant to say "without" the use of sections, rather than with :) Thanks for the responses so far, and I was more saying it in case it came off to the op that I was just being finicky. $\endgroup$ – snulty Jun 29 '16 at 18:17

A section of the bundles involved in the second and the third definition can be zero, and not define a metric.

After the change in your question, you can analyze the situation in a chart, (since metric are defined on charts then glued with partition of the unity) that is in an open subset $U$ of $R^n$, since the tangent bundle of an open subset of $R^n$ is trivial, consider the trivialization $(e_1,...,e_n)$

A metric $\langle,\rangle$ defined on $U$, induce a bilinear map on $TU$ of $b(e_i(x),e_j(x))=\langle e_i(x),e_j(x)\rangle$. The dual of $(e_1,...,e_n)$ defines a trivialization of $TM^*$ and to the metric $\langle,\rangle$ defined on $U$, you can associate the elements of $TM^*\otimes TM^*$ defined by $\sum c_{ij}(x)e_i^*(x)\otimes e_j^*(x)$ where $c_{ij}(x)=\langle e_i(x),e_j(x)\rangle$.

  • $\begingroup$ Edit; I clearly meant that those sections have to have the same properties as the first one, but just forgot to write it down. I think that's rather obvious, and that this does not answer or address the question at all. $\endgroup$ – B. Pasternak Jun 28 '16 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.