Push Forward on a Product Manifold.

[andresB]

Some words before the question.
For two smooth manifolds M and P It is true that
T(M\times P)\simeq TM\times TP
If I have local coordinates \lambda on M and q on P then (\lambda, q) are local coordinates on M\times P (right?). This means that in these local coordinates the tanget vectors are of the form a^{i}\frac{\partial}{\partial\lambda^{i}}+b^{i}\frac{\partial}{\partial q^{i}}

Now, I can compute push forwards in local coordinates. For example, for a function
f(\lambda, q)\rightarrow(\lambda,Q(q,\lambda))
Then
f^{*}\left(\frac{\partial}{\partial\lambda}\right)=\frac{\partial}{\partial\lambda}+\frac{\partial Q}{\partial\lambda}\frac{\partial}{\partial q}
where I just had to do the matrix product of the Jacobian to the column vector (1,0)^{T}.

Actual Question.

For a function f:\, TM\times TP\longrightarrow TM\times TP
and without using local coordinates what can be said about the Push forward f^{*}:\, TM\times TP\longrightarrow TM\times TP ?.

Particularly interested if the push forward can be descomposed into something in TM product something in TP.

 

[WWGD]

There is a result that for m in M , n in N, $T_{(m,n)} (M \times N) = T_m M (+) T_n N$ , where $(+)$ is the direct sum of (tangent) vectors. Where by '=' I mean isomorphic.

 

[andresB]

Thanks for replying.

Yes, I'm aware of the result, I actually implicitly used it in the example above.

 

[WWGD]

I am not sure of what you are looking for, but you can also use properties of duality, since $TM:= \cup (T_p M)^{*}$
Then $T(M\times N) = (T_{(m,n)} M \times N )^{*}=T_m^{*}M (+) T_n^{*}N = TM \times TN$.

Then a finite direct sum is a direct product .

You may want to play with these properties of duals , duals of products, etc. to look for the
result you want.

 

[andresB]

I'm not sure to understand what you mean. Either way, What I'm intersted is in the push forward.

If I have local coordinates, like in the example in the OP, for a given a function of M\times P to itself I can compute the push forward of that function. If I do not have coordinates then I have no idea what to do.

For example, let G be a lie group (I'm interested in SU(n)). The left translation is given by L_{a}b=ab. The push forward of the left translation aplied to a tangent vector at the identity, E\in T_{e}G , would give a tangent vector at the new group element (L_{a})_{*}E\in T_{a}G.

Now, for (g,g\text{´)}\in G\times G I would like to define L_{(1,h)}(g,g\text{´)}=(g,hg), and find the push forward of this function as L_{(1,h)*}(E\oplus0)= Something \oplus Something .