Manifold #11. The Cotangent Space매니폴드 #11. 코탄젠트 스페이스

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 11 of the playlist, The Cotangent Space. Up to the previous post we built the tangent space at a point $p$ of the manifold and its basis vectors. This time we build its dual space at the same point. It is where we see what the dual space from the Tensor series looks like on a manifold.

Recap of the previous posts

For each curve $\gamma$ through $p$ we made a velocity, and the collection of all these velocities was called the tangent space $T_pM$ at $p$. A velocity $Y$ acts on a smooth function $f \in C^\infty(M)$ and gives a real number.

$T_pM$ is a vector space, and once we choose a coordinate system, $\partial_1, \dots, \partial_d$ is a basis. From this post on I will drop the chart symbol and just say coordinates. We work within one coordinate system, and when we change coordinates we mark the new one with a prime, as in general relativity.

Covectors are made from functions

An element of the dual space was a linear map that takes a vector and gives a real number. So we need a map $\omega$ that takes a vector $Y$ of $T_pM$ and gives a real number. A vector of the tangent space has only one way to give a real number, which is to act on a function. So we fix one function $f$ and let every vector act on it. Just as a vector was attached to a curve, a covector is attached to a function.

The covector made from $f \in C^\infty(M)$ and the point $p$ is written $df_p$, and it is defined by

$df_p$ takes a vector $Y$ and returns the real number that comes from $Y$ acting on $f$. It is linear in $Y$ because $(Y + Z)(f) = Y(f) + Z(f)$.

Q. Can I understand it as a curve attached to a vector and a function attached to a covector?
A. Yes. A vector is the velocity of a curve through $p$, and a covector is the $df_p$ made from a function $f$. A real number comes out when a velocity $Y$ meets a function $f$, so for a covector to be a map that takes a vector and gives a real number, it needs a function.

The cotangent space

The collection of all such $df_p$ is called the cotangent space and written $T_p^*M$. The star is always there because it is the dual space of $T_pM$.

This collection is a vector space. The sum is $df_p + dg_p = d(f + g)_p$, and the scalar multiple is $r \, df_p = d(rf)_p$. It also has a zero element and additive inverses. The sum works because $d(f+g)_p(Y) = Y(f+g) = Y(f) + Y(g)$. (This check is added for understanding and was not covered in the lecture. The lecture states the result without proving that it is a vector space.) The dimension equals that of $T_pM$. The relation between the two is exactly $V$ and $V^*$ of the Tensor series: the tangent space is the underlying vector space $V$ and the cotangent space is its dual space $V^*$. The dual of the dual returns to the original space, so $T_pM$ is also the dual of $T_p^*M$. Now both spaces sit as a pair at every point of the manifold.

Q. Can different functions $f$ and $g$ give the same $df_p$?
A. They can. Adding a constant to $f$ does not change $Y(f)$, so $df_p$ stays the same. $df_p$ remembers not the whole of $f$ but only how $f$ changes to first order near $p$. (This explanation is added for understanding and was not covered in the lecture.)

The dual basis $dx^i$

The dual basis $\theta^i$ for the basis $\partial_j$ of the tangent space consists of covectors satisfying $\theta^i(\partial_j) = \delta^i{}_j$. The candidate is $dx^i$, made from the coordinate function $x^i$. A coordinate function is also a smooth function from the manifold to the real numbers, so we can form $dx^i$, and we compute with the definition above.

The coordinate $x^i$ depends on $x^j$ only when $i = j$, which gives the delta. In this computation we do not need to go back to the original definition that composes with the inverse of the chart. The notation was chosen to behave like partial derivatives, and this is exactly the reason for that choice. So any element $\omega$ of $T_p^*M$ is written with components $\omega_i$ as

In particular the components of a covector made from a function are $df_p = \dfrac{\partial f}{\partial x^i} dx^i_p$, the partial derivatives of the function. (That the components are partial derivatives follows at once by putting $\partial_i$ into the definition above, and it was not covered in the lecture.) It also follows that every covector has the form $df_p$ for some function. The $d$ of the function $\omega_i x^i$, the coordinate functions multiplied by the components and added, is $\omega$. Coordinate functions are defined only inside a chart, so strictly one uses a function trimmed to be 0 outside the chart. (This explanation was not covered in the lecture either.)

By the way, carrying $p$ around is a trace of the fact that we started from curves and had to fix the location of the point first. In the earlier Tensor series we did not write $p$, because we made one vector space and then said we would put a copy of it at every point. We can get rid of this $p$ when we deal with vector fields and enlarge the manifold into something bigger.

How do components change under a change of coordinates?

When the coordinates change from $x^i$ to $y^j$, the basis vectors change as $\partial_{y^j} = \dfrac{\partial x^i}{\partial y^j} \partial_{x^i}$. The vector $Y$ itself stays the same, so its components change in the opposite way.

The components of a covector are $\omega'_j = \omega(\partial_{y^j})$. Since $\omega$ is linear, we put in the transformation of the basis vectors as it is, and get

The $\partial y / \partial x$ multiplied into the vector components and the $\partial x / \partial y$ multiplied into the covector components are reciprocals of each other. This is the “opposite direction” relation the lecture mentions, and it is why vector components are called contravariant and covector components covariant. (The lecture skips the proof of this transformation and says only that it is nearly the same as the proof in the Tensor series. The derivation above is added to get it directly from the transformation of the basis vectors, and it was not covered in the lecture.) Because the two transformations are inverse to each other, $\omega'_j Y'^{\,j} = \omega_i Y^i$, so the real number a covector gives on a vector is the same in any coordinates. (This check is added for understanding and was not covered in the lecture.)

Q. Why are covector components called covariant?
A. The way a covector component $\omega_i$ changes uses the same derivative $\partial x / \partial y$ as the way the basis vector $\partial_i$ changes. It varies together with the basis, so it is covariant, and a vector component varies opposite to the basis, so it is contravariant. (This explanation is added for understanding and was not covered in the lecture.)

Try it in the toy below. Drag the point $p$ and the velocity $Y$, and change the function and the coordinates. The green lines are the level lines that $df_p$ makes, and the number of steps $Y$ crosses is $df_p(Y)$. When you change the coordinates, the components of $Y$ and of $df$ change, but the sum of the two stays the same.

Spaces stacked at every point, and bundles

Now that we have both the tangent and the cotangent space, we can also build every tensor product space made from them at the same point. These are all the spaces of type $(m, n)$ with $T_pM$ taken $m$ times and $T_p^*M$ taken $n$ times, and all of them are made from the curves through that point and the smooth functions used at that point. So each point of the manifold gets one copy of each of these spaces.

The next step is to build a bigger manifold that holds the manifold and these spaces all at once. The original manifold is called the base manifold. Attaching the tangent space of every point gives the tangent bundle, attaching the cotangent spaces gives the cotangent bundle, and attaching the tensor product spaces gives a tensor product space bundle. A bundle is itself a manifold with a topology, and it is locally Euclidean, so the dimension of the tangent bundle grows to $2d$, the $d$ of the base plus the $d$ of the tangent space. We deal with bundles in the next posts.