Manifold #10. Tangent Space - Basis Vectors매니폴드 #10. 접공간의 기저 벡터

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 10 of the playlist, Tangent Space - Basis Vectors. The earlier material is in the previous post.

What we got last time

The hard construction was finished in the previous post, so today is easier. We confirmed, by proving closure under addition, that the tangent space, which collects the velocities of all curves through a point $P$, is a vector space. The velocity of a curve $\alpha$ is a map that takes a smooth function $f$ and gives a real number like this.

Scalar multiples are closed in a similar way, so we may now call the elements of the tangent space vectors, not only velocities. A curve with the parameter changed from $t$ to $2t$ passes through the same point but moves twice as fast, so it is a different curve and its velocity is twice as large. Curves gather at every point, so every point of the manifold has its own tangent space. Even if one curve passes through two points, the tangent spaces at the two points are different vector spaces.

Today’s job is to build a basis of this vector space and to see how the components of a vector change when we change the coordinate system. The dual, the cotangent space, is left for the next lecture.

A basis needs one chart

When we computed the velocity of a curve, the chart made differentiating the curve possible. A basis likewise is defined only after we fix one chart. Pick one chart $(U, \gamma)$ from the atlas of the manifold $X$. Take a curve $\alpha$ passing through the point $P \in U$ at $t_0$, and an arbitrary test function $f$ from the manifold to the reals.

Insert the identity into the velocity

Written by definition, $v_{\alpha, t_0}(f) = (f \circ \alpha)'(t_0)$. Insert $\gamma^{-1} \circ \gamma$ here. It is the identity, so the value does not change.

The outer function $f \circ \gamma^{-1}$ goes from $\R^d$ to $\R$, and the inner function $\gamma \circ \alpha$ goes from $\R$ to $\R^d$. The multivariable chain rule gives a sum of products of partial derivatives and one-variable derivatives.

The first factor is the value of the $i$-th partial derivative of $f \circ \gamma^{-1}$ measured at $\gamma(P)$, and the second factor is the derivative in $t$ of the $i$-th coordinate of the chart representative of the curve, measured at $t_0$. The inner function has one variable, so it is multiplied as a plain derivative with no sum or index $i$ of its own. This formula is written by choosing the chart $\gamma$, so a different chart gives a different formula.

Components and basis vectors

Read the formula in two parts. One is a set of numbers determined only by the curve.

These are $d$ real numbers, the velocity of the chart representative $\gamma \circ \alpha$ written component by component. Changing the curve changes these numbers. The other is an operation that gives a number to $f$ regardless of the curve.

Here $x^i = \gamma^i$ is a coordinate function of the chart $\gamma$. This symbol is shorthand for the instruction to first compose $f$ with $\gamma^{-1}$ to make a function on $\R^d$, then take the partial derivative in the $i$-th variable and read the value at the chart representative of the point $P$. With it, the formula above becomes one line.

(We used the Einstein summation convention, which sums over an index appearing both up and down.) $X^i$ is a real number and $\partial/\partial x^i|_P$ is an operator acting on functions. So a velocity is a linear combination of numbers and differential operators.

Shortening the notation, and its dangers

This symbol is a shorthand that hides the chart. There is also a way to call the map itself $x$, writing the chart as $(U, x)$. Then $x(P) = (x^1(P), \dots, x^d(P))$, and another chart $(V, y)$ has $y(P) = (y^1(P), \dots, y^d(P))$. This notation, where the map and its coordinate functions share a letter, is widely used, but its weakness is that it is easy to confuse whether $x$ is one coordinate or the whole map.

Shortening hides two things. One is that $\partial/\partial x^i$ is the partial derivative not of $f$ itself but of $f \circ \gamma^{-1}$, a function from $\R^d$ to $\R$. This is because differentiation can be done only for that function. The other is that we write that the value is read at $P$, but it is really read at the chart representative $\gamma(P)$ of $P$. Still, this symbol is very efficient in calculations, and once you understand its meaning, you can regard $f$ as a function of the $x^i$ and just take partial derivatives.

Q. Why is $\partial / \partial x^i$ a vector? Is it not a differential operator?
A. Both are true. We defined the elements of the tangent space as maps that take a function and give a real number, and $\partial/\partial x^i|_P$ gives $f$ the real number $\partial_i(f \circ \gamma^{-1})(\gamma(P))$. This map is an element of the tangent space, so it is a vector, and its form of acting on functions makes it look like a differential operator.

The basis of the tangent space

We call $\partial/\partial x^1|_P, \dots, \partial/\partial x^d|_P$ the basis of $T_P X$ for the chart $\gamma$. That every vector of the tangent space has the form $X^i \, \partial/\partial x^i|_P$ is what we showed above, and this shape is the same as the basis we declared at each spacetime point in the Tensor series. What differs is that this time we produced this vector space from the definition of the manifold, and that it holds only at the one point $P$. When the point changes, the tangent space changes and so does the basis.

To be a basis, linear independence is also needed. Put the coordinate function $x^j$ itself in as the test function, and $\partial/\partial x^i|_P \, x^j = \partial_i(x^j \circ \gamma^{-1}) = \delta^j_i$, so if $a^i \, \partial/\partial x^i|_P = 0$ then $a^j = 0$ for every $j$. So the number of basis vectors is the dimension $d$ of the chart, and the dimension of the tangent space equals the dimension of the manifold. The same formula also reads off the components as $X^j = X(x^j)$. (This explanation is added to aid understanding and was not covered in the lecture.)

How do components change when the chart changes?

The basis is completely tied to the chart and its coordinate functions, because the test function $f$ also becomes a function of that coordinate system through $\gamma^{-1}$. Changing the coordinate system is changing the chart, so we compare two ways of writing the same vector in two charts. Let the coordinate functions of the chart $(U, \gamma)$ be $x^i$, those of the chart $(V, \varphi)$ be $y^j$, and suppose both contain $P$. The vector $X$ is written in two ways.

To express $X'^j$ through $X^i$ we must return to the definition. In the formula where the basis vector $\partial/\partial x^i|_P$ acts on a test function $f$, insert $\varphi^{-1} \circ \varphi$.

The inner part is the transition function $\varphi \circ \gamma^{-1} : \R^d \to \R^d$, and the outer part is $f \circ \varphi^{-1}$. By the chain rule, the $j$-th partial derivative of the outer function is multiplied by the $i$-th partial derivative of the $j$-th component of the transition function.

The first factor is $\partial/\partial y^j|_P f$, and the second is $\partial y^j / \partial x^i$, the partial derivative of the component $y^j$ of the transition function in the old coordinate $x^i$. This value is a known function, and it can be computed once the two charts are fixed. Erasing $f$ gives the transformation rule of the basis vectors.

Put this into $X = X^i \partial/\partial x^i|_P$ to get $X = X^i (\partial y^j/\partial x^i) \partial/\partial y^j|_P$, and comparing the coefficients of the same basis $\partial/\partial y^j|_P$ gives the transformation rule of the components.

The summed index is $i$ and the free index is $j$. The very transformation rule that general relativity textbooks use, with the meaning that new coordinates are functions of old coordinates, has come out of the definition of a manifold. Everything happens at the one point $P$, so $\partial y^j / \partial x^i$ is a number at that point.

Q. Where do we compute $\partial y^j / \partial x^i$?
A. We compute it at $\gamma(P)$, the old chart representative of $P$. It is the partial derivative of the $j$-th component of the transition function in the $i$-th variable, so the number changes when the point changes.

In the toy below, drag the point on the plane and the tangent vector $X$. The components of the same $X$ in the Cartesian basis $\partial/\partial x, \partial/\partial y$ and in the polar basis $\partial/\partial r, \partial/\partial \theta$ are shown side by side, and the values computed with the Jacobian matrix $\partial(r, \theta)/\partial(x, y)$ of the transformation rule rebuild the original vector. The value $X$ gives when acting on the test function $f = x^2 + xy$ is also the same in both charts. (This example with Cartesian and polar coordinates, and the toy, are added to aid understanding and were not covered in the lecture.)

Q. What happens to the components if we change the parameter of the curve to run twice as fast?
A. The curve $\alpha(2t)$ in place of $\alpha(t)$ is a different curve. The derivative of the $i$-th coordinate of the chart representative doubles, so every component $X^i$ doubles, and the vector is $2X$.

Summary and the next lecture

The basis vectors of the tangent space are the partial derivative operators $\partial/\partial x^i|_P$ with respect to the coordinate functions of a chart, defined by composing $f$ with $\gamma^{-1}$, taking the partial derivative on $\R^d$, and reading the value at $\gamma(P)$. When the chart changes, the basis changes by the Jacobian matrix $\partial y^j / \partial x^i$ and the components change by $X'^j = X^i \, \partial y^j / \partial x^i$. This is the same as the coordinate transformation rule of general relativity. We will not derive this rule again from now on and will go back to the form we used in the Tensor series. The next lecture covers the dual of the tangent space, the cotangent space.