# Pushforward of a vector

The push forward of vectors allows to transform the components of a vector $X$ to be pushed along a map $h:\mathcal{M}\to\mathcal{N}$ between the manifolds $\mathcal{M}$ and $\mathcal{N}$. This is done in a way, such that applied to a function $f$ on

$$f: \mathcal{N} \to \mathbb{R}$$

the following equality holds

$$X_* f = X \,(f \,\circ \, h)$$ where $X_*$ is the pushed vector along the map $h$ (notation is often $h_*X$).

As an example with one-dimensional manifolds take a chart ($\mathcal{M},x$) and on the other manifold ($\mathcal{N},y$) and define a vector $X = \cos(x)\frac{\partial}{\partial x^1}$ and the map between the manifold in these charts as $y = h(x) = x^2$. My goal is to to calculate $X_* X_*^1$, or in words I want to apply the pushed vector to its own components. This is an expression used to calculate the second application of a vector to a function. By definition of the push forward it can be found, that the pushed component is $$X_*^1 = \cfrac{\partial h}{\partial x} X^1 = 2x \cos(x) \quad .$$ Recall, that the components of a vector are functions on the manifold. So I can apply the vector $X_*$ to the first component $X_*^1$. And since $X_*$ was pushed to $\mathcal{TN}$, it can be written in the basis $\frac{\partial}{\partial y}$ which yields $$X_* X_*^1 = X_*^i \frac{\partial}{\partial y^i} X_*^1 = 2x \cos(x) \frac{\partial \, 2x\cos(x)}{\partial y}$$

and if I include the definition of the partial symbol, the last term is $$\frac{\partial \, 2x\cos(x)}{\partial y} = \partial_1 \left( 2x\cos(x) \,\circ\, y^{-1} \right) \quad .$$ That is strange, since I cannot derive the expression of "$x$ variables" with respepct to "$y$". Or do I just have to derive with respect to $x$, because this ?is? the first argument?

A possible solution to this simple example would be to invert the relation of $h$ and subsitute. This is not the solution I'm looking for, since in my real problem $h$ is a projection (or basically selects some components) and is therefore not invertible! I also know, that one can use the definition of the push forward and work with the original vector $X \in \mathcal{TM}$ and "move" the function $f$ by using $f \,\circ\, h$. However, in calculations some vector $X^i \frac{\partial}{\partial x^i} (f \,\circ\, h)$ always yields a chain rule, which could be avoided with the pushed vector. In fact, if one has to use the definition, what good is the pushed vector anyway!?

My real problem is the following: I have a dynamical system with a vector field $X$ on the mfd $\mathcal{M}$, where I have a global chart $(\mathcal{M},x)$. Also I got an "output equation" $y = h(x)$ in coordinates, which defines another manifold $\mathcal{N}$ and simultaneously a chart $(\mathcal{N},y)$ for this manifold. The map $h$ is a selection of states, which are of interest to me. So it's not a diffeomorphism. I now want to introduce new coordinates $(\mathcal{N},\xi)$ on $\mathcal{N}$. These are functions $$\xi : \mathcal{N} \to \mathbb{R}^p$$ Now I need the second Lie-derivative of one component function $\xi^a$ along the vector field $X$. Since the Lie-derivative of a function is just the application I write the thing I want as $$X_* X_* \xi^a = X_*^i \frac{\partial}{\partial y^i} (X_*^j) \frac{\partial}{\partial y^j} (\xi^a) + X_*^i X_*^j \frac{\partial^2}{\partial y^i y^j } \xi^a$$ by the product rule. What I was trying to simplify by using this $X_*$ is the term $$\frac{\partial}{\partial x^i} (X_*^j)$$ which is part of the first term. To make my life easy, I shortened $h_* X =: X_*$, which in components I can calculate to be $$X_*^i = h_{*a}^{i} X^a = \frac{\partial h}{\partial x^a} (y^i \circ h) X^a$$. And the whole vector is $$X_* = X_*^i \frac{\partial}{\partial y^i}$$. These components are so given in the $x$ chart and the above stated problems arise. The other way I can think of is using the above definition and keeping $(\xi \circ h)$ everywhere. But then I wonder why there is such a thing as push-forward, if I have to subsitute its definition all the time.

• What do you mean by $X = \cos (x)$ and by $h(x) = x^2$? May 23, 2016 at 12:22
• I tried to clarify that: $X$ is a vector (forgot the basis vectors to the components) on the maifold $\mathcal{M}$ and $h$ is a map between the manifolds $\mathcal{M}$ and $\mathcal{N}$.
– mike
May 23, 2016 at 16:48

I'm a bit confused by your notation: is $X_* f$ supposed to be a function on $\mathcal{M}$? Or on $\mathcal{N}$? Judging by the fact that you want $X_* = h_*X$ in the more usual notation, I assume the latter, in which case you want $$X_*f(y) = X[f\circ h]\left(h^{-1}(y)\right).$$

This gives you, in your example, $$X_*^1(y) = 2\sqrt{y}\cos(\sqrt{y}),$$ and now you can proceed with your calculation: \begin{align*} X_*^1 \circ h &= 2x\cos x\\ X(X_*^1\circ h) &= (\cos x)(2\cos x - 2x\sin x)\\ (X_*X_*^1)(y) &= \left(\cos \sqrt{y}\right)\left(2\cos\sqrt{y} - 2\sqrt{y}\sin\sqrt{y}\right). \end{align*}

• As already stated in the question: This is not the solution I'm looking for, since in my real problem h is a projection (or basically selects some components) and is therefore not invertible!
– mike
May 26, 2016 at 10:59
• @mike you will need to further explain what you mean by $X_*$, then, as currently it doesn't make sense / is ill-defined. $X(f\circ h)$ is a function $\mathcal{M}\to \mathbb{R}$, not $\mathcal{N}\to \mathbb{R}$. May 26, 2016 at 14:47
• Put differently, what do you expect to get if you evaluate $X_*f$ at a point that is not in the image of $f$? A point where the preimage contains multiple points on $\mathcal{M}$? You want to treat $X_*f$ as a function on $\mathcal{N}$, which only makes sense if $h$ is a diffeomorphishm. May 26, 2016 at 15:13
• I think I got hung up on my own example. I tried to explain where this problem arises, rather than focusing on the detail, after the line. To your question: I'd take $X_* f$ as a function on $\mathcal{N}$, yeah, since I apply a vector from $\mathcal{T}\mathcal{N}$ to a function on the mfd.
– mike
May 28, 2016 at 13:15
• After re-reading your solution, I assume the pushforwarded vector is only of use, if the map I'm pushing along is diffeomorphic?! Therefore, if it is, I can do, what you proposed, otherwise I have to live with the functions on $\mathcal{M}$, right?
– mike
May 28, 2016 at 14:55