# Defining a Riemannian metric

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 $\Gamma(T^\ast M)\otimes\Gamma(T^\ast M))\cong\mathrm{Hom}(\Gamma(TM)\otimes\Gamma(TM),C^\infty(M))$.

EDIT

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?

• Judging by your last sentence, is the $\Gamma$ in bullet point $3$ a typo? – snulty Jun 29 '16 at 16:25
• It's not. Is there anything wrong with it? – B. Pasternak Jun 29 '16 at 16:30
• 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 $\Bbb R$? – snulty Jun 29 '16 at 16:47
• There is a typo. Everything is bundles over $M$, but OP should have typed $\Gamma(\text{Hom}(TM\otimes TM,\Bbb R))$. – Ted Shifrin Jun 29 '16 at 16:53
• @TedShifrin You were right about the typo of course, I missed that because snulty asked about the $\Gamma$. – 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 $\Gamma(E\otimes F) \cong \Gamma(E)\otimes\Gamma(F)$.

• 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? – B. Pasternak Jun 29 '16 at 8:18
• 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. – Ted Shifrin Jun 29 '16 at 16:50
• @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? – snulty Jun 29 '16 at 17:20
• @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? – Ted Shifrin Jun 29 '16 at 17:44
• @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. – 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 $c_{ij}(x)=\langle e_i(x),e_j(x)\rangle$.

• 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. – B. Pasternak Jun 28 '16 at 20:14