Tensor #11. The Metric Tensor텐서 #11. 메트릭 텐서

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 11 of the playlist, The metric tensor. In the previous post we put an inner product on a vector space, and this time we show that this inner product is exactly the same as one map that already lives in a tensor product space.

Two kinds of maps

We have a 4-dimensional real vector space $\V$ and its dual space $\V^*$, with a basis $e_\mu$ and a dual basis $e^\mu$ tied together by $\map{e^\mu}{e_\nu} = \delta^\mu_\nu$. On $\V$ we put an inner product $\ip{v}{w}$, and it is fully fixed once we give 16 values on pairs of basis vectors.

Meanwhile the (0, 2) tensor product space $\V^* \otimes \V^*$ has elements of the form $T_{\alpha\beta}\, e^\alpha \otimes e^\beta$. Such an element is also a map that takes two vectors and returns a real number, and it likewise has 16 components. So there are now two ways to take two vectors and return a real number. One is the inner product put on $\V$, and the other is a tensor we get by choosing components in this space.

Choosing equal components

Can we choose, from the latter, something that behaves exactly like the inner product? We only need to set the components equal to the basis values of the inner product.

Put the vectors $a^\mu e_\mu$ and $b^\nu e_\nu$ into the inner product and, as in the last post, we get $a^\mu b^\nu \ip{e_\mu}{e_\nu} = a^\mu b^\nu g_{\mu\nu}$. Now put the same two vectors into the tensor $g_{\alpha\beta}\, e^\alpha \otimes e^\beta$. The map is linear, so all the constants come out.

The two values are exactly equal. This tensor, with the basis values as its components, gives the same real number as the inner product for any pair of vectors. We only added one inner product, yet a tensor that imitates its job exactly was already sitting inside $\V^* \otimes \V^*$.

The metric tensor

We call this tensor the metric tensor. It is the most important object in the basics of special and general relativity. As said in the last post, the inner product is a tool for measuring something, which is how the name metric arose. Put in two vectors and you get a real number comparing them, and put in the same vector twice and you get a real number corresponding to the squared size of that vector.

This tensor slightly breaks the promise to treat every element equally. The space has endless maps, and one of them gains a special status. But the direction of the gain can go either way. We can first put on an inner product and look for the tensor that imitates it, or conversely pull one map we like out of this space as the metric tensor and then fix an inner product that behaves exactly like it. Choosing one side makes the other follow immediately, so the two are completely the same. That is why calling $\V$ an inner product space and calling it a vector space with a metric mean the same thing. (Since the inner product is symmetric, $g_{\mu\nu} = g_{\nu\mu}$. This explanation follows from the requirement that the inner product be symmetric and was not covered in the lecture.)

Q. Doesn’t the metric tensor live in the dual space? You said the inner product is not there.
A. The inner product is put on the $\V$ side, so dual vectors cannot be compared with it. The metric tensor is an element of $\V^* \otimes \V^*$, a space built from the dual space. So what the inner product could not do opens up through this tensor.

Connecting vectors and covectors

Here the lecture steps back to point something out. We built only $\V$, and the dual space arose from logic alone. But there was no natural rule tying one vector of $\V$ to one covector of $\V^*$. That the basis vectors $e_\mu$ and $e^\mu$ look paired is only a rule we chose on purpose so that a delta function would appear. We can take the components of a vector as they are and build a covector, but that only says a covector with the same numbers exists, not that the two are related in any way.

With the metric tensor this changes. Give it a vector, and the covector that goes with it comes straight out of the metric tensor.

Filling only one slot

Put the vector $a^\alpha e_\alpha$ only into the first slot of the metric tensor and leave the second slot empty. This map waits for two vectors and received only one, so it cannot return a real number, and something that still waits for a vector remains.

The first slot disappeared as the delta function turned $\alpha$ into $\mu$, and the second slot remained as $e^\nu$. The remaining $e^\nu$ is a basis of covectors, which return a real number when given one vector. So this is an element of $\V^*$ with components $a^\mu g_{\mu\nu}$. The upper index $\mu$ was contracted with the lower index of $g$ and vanished, so only the free index $\nu$ is left. This happens so often that we agree to drop the contracted index from the name altogether.

The agreement is that an $a$ with a lower index $\nu$ is “what we get by contracting the upper-index $a^\mu$ with the metric tensor”. This process is called lowering an index, and it cannot be done without the metric tensor. Calling the upper-index components $a^\mu$ contravariant components and the lower-index components $a_\nu$ covariant components now follows as well. There are two names, but both come from one vector, and as explained before they are components that change differently under a change of basis. Still, $a_\nu e^\nu$ is a covector living in $\V^*$ and $a^\mu e_\mu$ is a vector of $\V$. The names are similar, but they live in different spaces, and the metric tensor is what connects them.

(Because of symmetry, $a^\mu g_{\mu\nu} = g_{\nu\mu}\, a^\mu$, and whichever slot we fill first gives the same covector. This explanation follows from the symmetry of the metric tensor and was not covered in the lecture.)

Q. What comes out if I put the vector $b^\nu$ into the lowered $a_\nu$?
A. We filled the remaining slot with a vector, so a real number comes out, and it equals $g_{\mu\nu} a^\mu b^\nu$, where both slots were filled from the start. That is $\ip{a}{b}$.

Lowering one yourself

Let us lower one in 2 dimensions. If you pick the Minkowski metric, $a_\nu$ and $a^\nu$, which come from the same vector, have different numbers. The sign of the time component flips. (This explanation is added to give an example and was not covered in the lecture.) Conversely, if $g$ is the identity the numbers are equal, but even then the two are elements of different spaces.

Q. Can we get back from the lowered covector to the original vector?
A. We can if the determinant of the metric tensor is not 0. Then we multiply by the inverse of $g$ to recover $a^\mu$. This post does not cover the way back. (This explanation is a fact you can check in the toy and was not covered in the lecture.)

To higher orders

This process does not stop at vectors. For higher-order tensors whose components have several upper indices, we can lower the indices one at a time in the same way. In the next post we look at an example briefly.