# Notes on Metric Tensors

These are some notes starting after Nicolas Behr’s physics talk at the School of Informatics on Tuesday 8th july 2014. It is very hand-wavy but hopefully might give a useful insight into metric tensors

Afterwards, I asked what the difference between an outer product and a tensor product is, and wrote on the board something that looked like high-school linear algebra

\begin{equation} \left( \begin{array}{c} 1 \\ 2 \end{array} \right) \left( \begin{array}{cc} 3 & 4 \end{array} \right) = \left( \begin{array}{cc} 3 & 4 \\ 6 & 8 \end{array} \right) \end{equation}or writing down the same thing in a slightly better way,

\begin{equation} \vec{u} \otimes \vec{v} = u_a v^b = w_a^b \end{equation}which Nicolas corrected to include the basis vectors, but basically agreed an outer product like this, which is just all the different ways we can multiply the components (in the forward direction) is basically equivalent to the tensor product as long as we can keep track of which elements go where.

\begin{equation} \vec{u} \cdot \vec{v} = \sum_a u_a v^a = \sum_a w_a^a = Tr(w) \end{equation}As usual if we follow Einstein’s summation convention, we should stop writing the big sigma because we quickly end up with too many of them. But for the time being we’ll keep them to make that obvious.

The indices are written a bit randomly. Why should they be up or down? The answer here is that for the dot product it doesn’t matter. The trace of \(w\) is invariant under (linear) coordinate transformations so getting confused about which way the rows and columns go doesn’t make a difference.

Obviously it gets harder and harder to write down as the rank goes up – I have no idea how to write down a rank-4 tensor in matrix form which is what you’d get doing the outer product of that with itself.

So then an inner product is what you get when you contract an outer product where “contract” means taking the trace (for rank 2, it would be something analogous but harder to visualise for higher rank).

With elementary linear algebra we also have a cross product, which is where this stuff gets much harder to write down with rank > 2. But it seems natural that we should be able to generalise the idea of cross products to these things. It turns out that we need a new symbol to help us do this, a permutation symbol named after Levi-Civita (who was a mathematician that wrote one of the main early works in this area, the somewhat starkly titled “The Absolute Differential Calculus”).</p>

Anyways, this symbol \(\epsilon_{ijkl…}\) is 1 if the indices are an even permutation of \((1,2,3,4\ldots)\), -1 if it is odd, and zero if any indices are repeated. With this thing,

\begin{equation} \vec{u} \times \vec{v} = \sum_j \sum_k \epsilon_{ijk} u^j v^k \end{equation}You can check that this works directly for the simple cases of, say, three dimensional vectors in euclidean space that we are all familiar with, and can just prove by induction that it works in higher dimensions.

So far all of this is just a different way of writing down linear algebra. Things get a bit hairy when we don’t have a linear basis. If we have spherical coordinates, trying to do dot or cross products by multiplying r’s and thetas together without paying attention clearly won’t work. We need a thing to tell us how to do that.

The important quantity that we want to preserve when we do these things is distance. The length of a vector, which we get by using dot products, should be the same no matter what coordinate system we use. The thing that does this, that tells us how to multiply vectors together in such a way that lengths are preserved, how much of each coordinate to put into the result, is the metric tensor. It’s called “metric” for just this reason.

We know how to compute lengths in euclidean spaces, we know how to (locally) transform from euclidean space to some sort of curvy space, so we can work backwards. We can start by writing down our dot product as

\begin{equation} \vec{u} \cdot \vec{v} = \sum_i \sum_j \delta_{ij} u^i v^j \end{equation}Nothing has really changed here, the delta is just our normal Kronecker delta that is 1 if \(i = j\) and 0 otherwise. But what if it wasn’t? We know more about our metric tensor now. We know it fits in the same place as the delta. So let’s go ahead and write down something similar, where everything is expressed in some other curvilinear coordinate system

\begin{equation} \vec{n} \cdot \vec{m} = \sum_i\sum_j g_{ij}n^im^j \end{equation}we want that they are the “same” vectors just represented with different coordinates, so we can write,

\begin{equation} \begin{aligned} n^i &= \sum_j \frac{\partial y^i}{\partial x^j} u^j\\ m^i &= \sum_j \frac{\partial y^i}{\partial x^j} v^j \end{aligned} \end{equation}where \(x^i\) are the old rectilinear coordinates and \(y^i\) are the new curvilinear ones. We want to make sure that

\begin{equation} \vec{u} \cdot \vec{v} = \vec{n} \cdot \vec{m} \end{equation}so,

\begin{equation} \sum_i\sum_j g_{ij}n^im^j = \sum_i\sum_j\sum_k\sum_l g_{ij} \frac{\partial y^i}{\partial x^k}u^k \frac{\partial y^j}{\partial x^l}v^l = \sum_i\sum_j \delta_{ij} u^iv^j \end{equation}this is a bit of a beast with lots of sums, but rearranging isn’t that hard, and we get that

\begin{equation} g_{ij} = \sum_a\sum_b \frac{\partial x^a}{\partial y^i} \frac{\partial x^b}{\partial y^j} \delta_{ab} \end{equation}In a similar way we do cross products,

\begin{equation} \vec{n} \times \vec{m} = \sum_j \sum_k \sum_l g_{ij} \epsilon_{jlk} n^k m^l \end{equation}The effect of applying a metric tensor this way (“applying” means summing over its second index) is a kind of a distance preserving coordinate transformation.

Whether we put the indices on the top or the bottom depends on whether coordinate transformations happens in a covariant (top) or contravariant (bottom) way. Normally vectors are covariant things. If you make the vector bigger in some direction, you would expect its representation in another coordinate system also to get bigger.

The metric tensor is the opposite. It’s supposed to make lengths (dot products) stay the same no matter how you transform. So if you go to a coordinate system that stretches in the \(x\) direction, assigns a bigger number to that component, the metric tensor has to compensate and assign a smaller number to the bit of the length sum that comes from the corresponding components in the transformed vector. So we call it contravariant.

A historical note is that this mathematics came to physics with general relativity. Special relativity can be worked out with high school algebra, but in general relativity, there was a distance quantity that needed to be conserved under acceleration and there was no way to make it work with euclidean coordinates. So the idea was that it could be worked out with curvilinear coordinates and that this metric tensor arranges so that the distances stay the same.

It’s also hopefully clear why it’s good to use the summation convention and drop all of the annoying sigmas.