# Matrix Multiplication on Riemannian Manifolds

I am having a hard time understanding the concept of matrix (and / or vector) multiplication on a Riemannian Manifold $$(M, g)$$.

On $$\mathbb R^n$$ we can multiply a matrix for a vector in the usual way. How do I translate that on $$M$$? The naive way would be to just do the multiplication on the local coordinates, but this entirely disregards the metric, which seems wrong.

Is the matrix multiplication something that lives on $$T_vM$$? Intuitively yes, but why?

For a general Riemannian Manifold $$(M,g)$$ it isn't clear what your question means. There is not a canonical way to let a matrix act on an element of the manifold $$M$$ itself. On the other hand, if we fix a point $$p\in M$$, then $$T_pM$$ is a vector space of (say) dimension $$n$$. In this case, if $$A\in \mathcal{M}_n(\mathbb{R})$$ we can act on elements of $$T_pM$$ by providing a basis and using left multiplication. Take a coordinate system $$(U,x^1,\ldots, x^n)$$ near $$p$$. There is an associated vector field $$(\partial_1,\ldots, \partial_n)$$ on $$U$$, so that $$(\partial_1|_p,\ldots,\partial_n|_p)$$ forms a basis at $$T_pM$$. If we identify vectors in $$T_pM$$ with arrays of real numbers in this basis, $$A$$ acts by multiplication.
• If $M$ has a vector space structure then most likely you want $M$ to be equipped with a (global) inner product $\langle\cdot,\cdot\rangle$ which provides the Riemannian structure. That is, $M$ should be an inner product space. Once we have this and once we fix a basis $(e_1,\ldots, e_n)$ for $M$, a matrix $A$ acts as usual: by left multiplication of the vectors. Actually, inner product spaces are the archetypal version of Riemannian manifolds. Commented Feb 4, 2019 at 22:28
• How does the inner product provide the Riemannin structure? $g(v, w) = <v,w>$?
• Yes, same inner product for all $p\in V$. Commented Feb 6, 2019 at 13:06