Why a metric tensor is considered a tensor as opposed to a matrix?

Question

My understanding is that a metric tensor receives two points and returns a distance. Why is it considered a tensor then? (Tensor typically means a high order matrix that needs more than two index for each entry.)

Answer 1

A matrix is a bunch of numbers arranged in a rectangular grid. A thing that receives two [tangent vectors] and returns a number is not that.

A bilinear functional is not a matrix, for the same reason that a linear map is not a matrix. For a simple example, consider the rotation of a plane (2-dimensional Euclidean space) by 90 degrees. Is this a matrix? No. It may be represented by a matrix after a basis is chosen in the space. (And the matrix will depend on that choice.) Until then, it has no such representation.

In a Euclidean space, we get to pick a basis early in the game, and from that point on, happily mix linear maps, matrices, and bilinear forms. But on a general manifold, there isn't a canonical basis for a tangent space at every point. So we have to play the game without one, and consequently, without a matrix representation of the metric tensor.

Answer 2

A linear map is just an element of $\operatorname{Hom}_{\Bbbk}(W,V)$, where $V,W$ are $\Bbbk$-vector spaces; using the dual pairing this can be identified with $V\otimes_\Bbbk W^*$. Once you have chosen bases on $V$ and $W$ you can represent a linear map $L\in V\otimes_\Bbbk W^*$ as a $\dim W\times \dim V$ matrix with coefficients in $\Bbbk$.

Now, since a second rank tensor field at a point is just an element of \begin{equation} \begin{array}{ll} V\otimes_{\Bbbk} V\simeq\operatorname{Hom}_\Bbbk(V^*,V) & \text{rank}=(2,0)\\ V\otimes_\Bbbk V^*\simeq\operatorname{Hom}_\Bbbk(V,V) &\text{rank}=(1,1)\\ V^*\otimes_\Bbbk V^*\simeq\operatorname{Hom}_\Bbbk(V,V^*) &\text{rank}=(0,2) \end{array} \end{equation} where $V=T_xM$ is the tangent space at $x\in M$ (which is a $\dim M$ $\Bbbk$-vector space), and $\Bbbk=\mathbb{R},\mathbb{C}$, once you have picked local bases on $V$ and its dual you can represent it with a matrix.

Moreover, if the manifold has a metric, there is an isomorphism between $V$ and $V^*$, called the musical isomorphism or, by physicists, raising/lowering the indices, which lets you relate all the $r$-rank tensors by concracting with the metric tensor.

For example, consider the Minkowski metric (which is not a Riemannian metric, but a Lorentzian one) on $\mathbb{R}^{1,3}$; with respect to the canonical basis $\lbrace e_i\rbrace$ of $V$ (and its dual $\lbrace e^i\rbrace$ in $V^*$) \begin{equation} \eta(e_i,e_j)=:\eta_{ij}=\begin{bmatrix} -1 & 0 & 0& 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 0 &0 &1 \end{bmatrix},\quad \eta=\sum_{ij}\eta_{ij} e^i\otimes e^j \end{equation} since $\eta$ is invertible, its inverse is an element of $V\otimes_\Bbbk V$ (it is usually denoted with the same symbol) and can be represented by the matrix \begin{equation} \eta(e^i,e^j)=:\eta^{ij}=\begin{bmatrix} -1 & 0 & 0& 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 0 &0 &1 \end{bmatrix},\quad \eta=\sum_{ij}\eta^{ij} e_i\otimes e_j \end{equation} and therefore \begin{equation} \sum_{j}\eta^{ij}\eta_{jk}={\eta^i}_k=\delta^i_k,\quad \eta=\sum_{ij}{\eta^i}_j e_i\otimes e^j. \end{equation}

Finally, I'd like to stress again that a metric is more than a tensor, it is a tensor field, i.e. a $\mathcal{C}^\infty_M$-map which assigns to each point $x\in M$ a rank $(0,2)$ tensor which induces an inner product (in the case $\Bbbk=\mathbb{R}$) on each tangent space.