1. Apr 6, 2016

### dyn

Hi. For a diffeomorphism between 2 manifolds Φ : M → N with a tangent vector v in M I have the following equation for the push-forward of v
( Φ*v)f = v( Φ*f) where Φ* is the pull-back. I understand this equation but i have also come across the following equation for the push-forward
*v)f = (Φ-1)*v (Φ*f ) . Surely these 2 equations are not the same. I'm confused.

2. Apr 6, 2016

### andrewkirk

I assume that $v$ is a vector in a tangent space of $M$ and $f$ is a one-form in a cotangent space of $N$.
Then the RHS of the second equation doesn't seem to make any sense. $(\Phi^{-1})^*$ needs as argument a one-form in the relevant cotangent space of $M$, but what it is given $v(\Phi^*f)$, which is a scalar. So the RHS is undefined - meaningless.

Where did you see that second formula? Perhaps it is a typo.

3. Apr 6, 2016

### Orodruin

Staff Emeritus
You can use the pullback of a one-form to define the pullback of an arbitrary p-form. For a 0-form this is rather uninteresting though so maybe not what was intended.

4. Apr 7, 2016

### Ben Niehoff

I'm fairly sure what's intended is that $v$ is a vector on $M$ and $f$ is a function on $N$. The action of $v$ on $f$ (if they were both on $M$, which they are not) is defined by

$$v(f) \equiv df (v)$$
(or alternatively, that's the definition of $df$, depending on which notions you've decided are more fundamental).

If $f$ is a function on $N$ given by $f : N \to \mathbb{R} ; y \mapsto f(y)$ for $y \in N$, then the pullback $\Phi^* f : M \to \mathbb{R}$ is defined, for $x \in M$, via

$$(\Phi^* f)(x) \equiv f(\Phi(x))$$
Now, as for your confusion about the equations. $\Phi_* v$ should be a vector field living on $N$, and hence $(\Phi_* v)(f)$ should be a function living on $N$. However, in your first equation, $v(\Phi^* f)$ is clearly a function on $M$, not $N$. Therefore, it is your second equation which is correct:

$$(\Phi_* v)(f) \equiv (\Phi^{-1})^* (v(\Phi^* f))$$
because now both sides of the equation live on $N$. One must use $(\Phi^{-1})^*$ rather than $\Phi_*$, because functions out of a space (as $f$ is) must be pulled back rather than pushed forward.

An interesting question arises when perhaps $\Phi^{-1}$ doesn't exist (for example, when $M$ has smaller dimension than $N$, and $\Phi$ is an embedding). In this case $\Phi_* v$ is not defined on all of $N$, but only on the portion of $N$ on which $\Phi$ is invertible. That is, $\Phi_* v$ is only defined on the image of $\Phi$.

5. Apr 7, 2016

### andrewkirk

I suppose it depends on whether $v$ is a vector field or just a single vector in a single tangent space at point $p\in M$. From the way the question is worded - not mentioning vector fields - I feel drawn to assume the latter.

In that case $v(\Phi^* f)$ is a scalar in the overarching field $F$, being the directional derivative in direction $v$ of the scalar function $(\Phi^* f):M\to F$, at point $p$. The equation then asserts that that is equal to $(\Phi_*v)f$, which is the directional derivative in direction $\Phi_*v$ of the scalar function $f:N\to F$, at point $\Phi(p)\in N$.

I don't know whether that equation is valid, but it is well-defined as an equality between two elements of the same field.

I think more context of the problem is needed to make a clear interpretation.

6. Apr 7, 2016

### dyn

The vector v is a tangent vector in the manifold M. Of the 2 equations I quoted does one refer to a function in M and one refer to a function in N ? Do functions exist only in specific manifolds ?

7. Apr 7, 2016

### dyn

I forgot to say ; thanks for all your replies.

8. Apr 7, 2016

### andrewkirk

Assuming that $f$ represents a function and not a one-form in this context, which seems likely, then both equations refer to both a function $f$ on $N$ and a function $(\Phi^*f)$ on $M$. The latter is the 'pullback' of the former and, as Ben pointed out above, is defined by

$$(\Phi^*f)(p)=f(\Phi(p))$$

where $p\in M$.