Tensor #13. Realization of a Vector Space텐서 #13. 벡터 스페이스의 리얼라이제이션

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 13 of the playlist, Realization of a Vector Space. In the previous post we learned the rules for raising and lowering indices, so I first use those rules to build the inner product of the dual space, and then turn to the realization, which gives concrete objects to the basis vectors that were only symbols.

The inner product of the dual space

Once an inner product is given on the vector space $\V$, the metric tensor $g_{\mu\nu}$ follows. The metric pairs each vector with one covector.

With the pairing fixed, the inner product of two covectors can be defined as the inner product of the vectors they are paired with. In other words, there is nothing new to choose. Writing the inner product of the dual space with a star to tell it apart, the inner product of basis covectors comes out like this.

Here $g^{\mu\nu}$ is the inverse matrix of $g_{\mu\nu}$. (This derivation adds the standard definition and was not covered in the lecture. The vector components paired with covectors $\alpha_\mu$ and $\beta_\nu$ are $g^{\mu\rho}\alpha_\rho$ and so on, and simplifying their inner product $g_{\mu\nu}\, g^{\mu\rho}\alpha_\rho\, g^{\nu\sigma}\beta_\sigma$ gives $g^{\rho\sigma}\alpha_\rho\beta_\sigma$.)

Then there must also be a metric tensor for this inner product. It is a map that takes two covectors, so it is an element of $\V \otimes \V$, that is, of type (2, 0), and we write it $g^{\mu\nu} e_\mu \otimes e_\nu$. The metric $g_{\mu\nu} e^\mu \otimes e^\nu$ with lowered components is of type (0, 2), so the three shapes of the metric live in different spaces. The mixed shape $g^\mu{}_\nu$ is the identity map $\delta^\mu{}_\nu$, which lives in $\V \otimes \V^*$. (This explanation comes from working out the result of raising an index and was not covered in the lecture.)

Q. Do we have to define the inner product of the dual space separately?
A. There is no need. In the same way that the dual space and the tensor product spaces came for free when we built the vector space, giving an inner product brings the metric and the inner product of the dual space along with it.

A basis that was only symbols

So far the vector space has been an abstract object put in one box. It is four-dimensional over the real numbers and its basis is $e_0, e_1, e_2, e_3$. These $e_\mu$ are only typographical symbols. The rules held for them because we defined their sums and real multiples to follow the vector space rules. This is a different route from a concrete object, such as an arrow in physics, that satisfies the rules by itself and therefore becomes a vector.

A realization means finding real mathematical objects that satisfy these rules, namely addition, real multiples, and the zero and opposite vectors, all closed, and using them in place of the symbols. The most important candidate is the differential operator.

The space of differential operators

Take a differentiable function $f(x^0, x^1, x^2, x^3)$. We write the partial derivative with respect to each coordinate as $\partial_\mu f = \partial f / \partial x^\mu$. The upper index in the denominator comes down as a lower index, so the Einstein summation convention works as it does in $\partial_\mu f\, a^\mu$.

Differentiation is linear. We have $\partial_0 f + \partial_2 f = (\partial_0 + \partial_2) f$, and for real numbers $a, b$ we have $a\,\partial_0 f + b\,\partial_2 f = (a\,\partial_0 + b\,\partial_2) f$. So the operators themselves can be added and multiplied by real numbers, and a general operator is written like this.

The sum of two operators is the sum of their coefficients. We have $(a\,\partial_0 + b\,\partial_1) + (c\,\partial_0 + d\,\partial_1) = (a+c)\,\partial_0 + (b+d)\,\partial_1$, and there are a zero operator and an opposite operator with $L + (-L) = 0$. The set of all these operators is a four-dimensional vector space over the reals that obeys every rule of a vector space, and $\partial_\mu$ is a basis. Putting $\partial_\mu$ in the place of the symbols $e_\mu$ is the realization.

The function $f$ is not important here. We only add the operators and multiply them by real numbers, and we are not computing any derivative. Having already learned to handle maps as objects gathered together, we can see a toolbox of tools for functions becoming a vector space.

Q. Are $\partial_0, \dots, \partial_3$ really a basis? They must be independent.
A. Apply $L = a^\mu\partial_\mu$ to the coordinate function $x^\nu$ and we get $L\, x^\nu = a^\nu$. If $L$ is the zero operator then every $a^\nu$ must be 0, so they are independent. (This explanation checks the independence and was not covered in the lecture.)

One vector space at each point

Now to spacetime. One way to name every point of spacetime is to give four coordinates $x^0, x^1, x^2, x^3$, abbreviated $x^\mu$. This upper index does not mean $x$ is an element of a vector space. If it were a vector, a basis would sit beside it, so here the index only marks that there are four coordinates. Once we have coordinates we can make as many functions from spacetime to the reals as we like, and there is something to differentiate.

Here we change how we think about the vector space so far. We discard the abstract basis $e_\mu$ and replace it with $\partial_\mu$, and we place one vector space at every point of spacetime. The one placed at a point P is written $\V_P$ and the one at a point Q is $\V_Q$. These spaces have no relation to one another, and only their names are decided by the coordinates of the points.

A vector of $\V_P$ and a vector of $\V_Q$ cannot be added, because only vectors living in the same vector space can be added. Seen as operators the reason is clearer: $\partial_\mu$ means taking the derivative at P, which is a different operator from the derivative taken at Q. So the correct way is to attach the point, as in $A^\mu \partial_\mu|_P$. In the toy below, drag the points P and Q and change where $L_2$ sits to check this. The picture is a plane with two coordinates.

Q. The components $A^\mu$ and $B^\mu$ are just numbers. Why can they not be added?
A. The numbers can be added, but the result is not a vector at any point. A vector is an object that carries components together with a basis, and the bases of the two are operators at different points, so there is nowhere to put the sum.

The structure that comes with each point

Once there is a vector space, the dual space and infinitely many tensor product spaces follow. So each point carries all of them. There are uncountably many points, so the amount of structure goes beyond imagination. If we say there is a metric at every point, each of these spaces also gets an inner product. How to bundle these inner products and handle them together is the heart of general relativity, and it is what differential geometry does.

One point to note. What we saw just now is only the skeleton that puts a vector space at each point. We have not yet said anything about choosing one vector at each point.

In the next post we move to the concept of a field, which handles all these spaces at once.