Cotangent space and pushforward

co-tangent-spacedifferential-geometrypushforward

I wonder if there is a connection between the following two notions:

Pushforward: For a smooth map $f:M\to N$ between smooth manifolds $M$ and $N$, we define the pushforward
$$df:TM\to TN$$
between the tangent spaces $TM$ and $TN$. For a derivation $X\in T_xM$, we define $df(X)=df\vert_x(X)$ to send a smooth map $g:N\to\mathbb R$ to $X(g\circ f)$.

Cotangent Space: On the other hand, we also define the cotangent space of $M$ at $x\in M$ to be $T^*_xM$, the dual space of $T_xM$. If we have a chart $\phi=(x_1,\dots,x_n)$ around $x$, we have a basis $\frac{\partial}{\partial x_i}$ with $1\leq i\leq n$. It induces a dual basis on $T^*_xM$, given by the linear maps $dx_i\vert _x:T_xM\to\mathbb R$, defined by $dx_i\vert_x\big(\frac{\partial}{\partial x_j}\big)=\delta_{ij}$ (Kronecker delta).

Why do we use the same notation for both? If I consider for instance the map $$x_i:\mathbb R^n\ni(x_1,\dots,x_n)\mapsto x_i\in\mathbb R,$$
it is unclear if by $dx_i\vert_x$ I mean the pushforward map or an element of the dual basis of $T_x^*\mathbb R^n$. I figured both cannot be the same, though:

  • If we consider $dx_i\vert_x$ as a pushforward, then $dx_i\vert_x(X)$ is a derivtion (hence a map).
  • If we consider $dx_i\vert_x$ as a basis element of $T_x^*\mathbb R^n$, then $dx_i\vert_x(X)$ is a real number.

While writing this, I got the idea that the derivation $dx_i\vert_x(X)$ is just an element of a one-dimensional vector space with a canonical basis, so we can just identify it with the corresponding scalar. Is this the connection between both?

Best Answer

There is indeed a relationship between the pushforward and the differential. This relationship goes further than what you describe: it applies to all smooth functions $f : M \to \mathbb R$, not just to the coordinate functions $x^i$.

Can I suggest we use different notation for the pushforward and the differential, so that we don't get confused?

Let $f : M \to N$ be a smooth map and let $x \in M$. The pushforward $(Df)_x$ is a linear map $$(Df)_x : T_x M \to T_{f(x)}N.$$Applying $(Df)_x$ to a tangent vector $X \in T_x M$, we get a tangent vector $(Df)_x(X) \in T_{f(x)}N$. The action of $(Df)_x(X) \in T_{f(x)}N$ on any smooth function $g : N \to \mathbb R$ is defined to be $$(Df)_x(X) (g) = X(g \circ f). $$

Let $f : M \to \mathbb R$ be a smooth function and let $x \in M$. The differential $(df)_x$ is an element of the cotangent space $(T_x M)^\star$; that is, it is a linear map $$(df)_x : T_x M \to \mathbb R.$$ Applying $(df)_x$ to a tangent vector $X \in T_x M$, we get the number, $$ (df)_x(X) = X(f).$$

In your question, you considered some local coordinates $\{ x^i \}$ around the point $x$, and you described how the $(dx^i)_x$'s form a basis for the cotangent space at $x$. These $(dx^i)_x$'s are precisely the differentials of the local coordinates $x^i$. As we'll soon see, these differentials $(dx^i)_x$ are related to the pushforwards $(Dx^i)_x$. (Okay, strictly speaking, my definitions for the pushforward $(Df)_x$ and differential $(df)_x$ require the smooth function $f$ to be globally defined on $M$, whereas the $x^i$'s are only locally defined around $x$. However, you can use a partition of unity to construct for each $x^i$ a smooth function $\widetilde{x^i}$ defined on the whole of $M$ which agrees with $x^i$ on a neighbourhood of $x$. You can then legitimately compute the pushforward $(D\widetilde{x^i})_x$ and differential $(d\widetilde{x^i})_x$, and the answers you get are independent of your choice of $\widetilde{x^i}$.)

Anyway, let's return to comparing the notions of the pushforward and the differential. For any smooth function $f : M \to \mathbb R$:

  • Taking the manifold $N$ to be $\mathbb R$, the pushforward $(Df)_x$ is a linear map $T_x M \to T_{f(x)} \mathbb R$.
  • Meanwhile, the differential $(df)_x$ is a linear map $T_x M \to \mathbb R$.

It looks like we're almost there. We just need to identify $T_{f(x)} \mathbb R$ with $\mathbb R$, somehow.

As you observed in your original post, both $T_{f(x)} \mathbb R$ and $\mathbb R$ are one-dimensional vector spaces. Moreover, there is a canonical vector space isomorphism $$ \vartheta_{f(x)} : T_{f(x)} \mathbb R \overset{\cong}{\to} \mathbb R.$$ $\vartheta_{f(x)}$ is defined as follows. Given a tangent vector $Y \in T_{f(x)} \mathbb R$, $\vartheta_{f(x)}$ maps $Y$ to the number $$ \vartheta_{f(x)} (Y) := Y(\text{id})$$ where $\text{id} : \mathbb R \to \mathbb R$ is the identity map on $\mathbb R$.

It should be easy to see that the relationship $$ \vartheta_{f(x)} \circ (Df)_x = (df)_x $$ holds for all smooth functions $f : M \to \mathbb R$ and for all points $x \in M$. This is the relationship between $(Df)_x$ and $(df)_x$ that you were looking for.

To make things a little more tangible, let's run through this again, this time using some local coordinate systems.

  • Suppose that $\mathbf x : = (x^1, \dots, x^n)$ are local coordinates on a neighbourhood of $x$ on $M$. Suppose that the coordinates of our point $x$ are $\mathbf x = \mathbf x_0$.
  • Suppose that $t$ is the canonical coordinate for $\mathbb R$. Suppose that $f(x)$ is the point $t = t_0$.
  • Suppose that $\hat f$ is the coordinate representation of our smooth map $f$ with respect to our $(x^1, \dots, x^n)$ and $t$ coordinates.

A general tangent vector $X \in T_{x} \mathbb R$, takes the form $$ X = \sum_{i = 1}^n c^i \left. \frac{\partial}{\partial x^i} \right|_{\mathbf x_0},$$ where $c_i$ are numbers in $\mathbb R$.

We have $$ (Df)_x (X) = \sum_{i = 1}^n c^i \frac{\partial \hat f}{\partial x^i}(\mathbf x_0) \left. \frac{d}{dt} \right|_{t_0}$$ and $$ (df)_x(X) = \sum_{i = 1}^n c^i \frac{\partial \hat f}{\partial x^i}(\mathbf x_0).$$

The canonical isomorphism $\vartheta_{f(x)} : T_{f(x)} \mathbb R \to \mathbb R$ is given by $$ \vartheta_{f(x)} \left( a \left. \frac{d}{dt}\right|_{t_0} \right) = a, $$ where $a$ is any number in $\mathbb R$.

Thus it's evident that $\vartheta_{f(x)}((Df)_x(X)) = (df)_x(X)$.

Related Question