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 $f:\, M\times P\longrightarrow M\times P$ 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 decomposed into something in $TM$ plus something in $TP$

  • 1
    $\begingroup$ The answer to your first and second question is yes. Also, the pushforward is usually denoted $f_\ast$ not $f^\ast$, this is the pullback. Also I think you mean $f:M\times P\to M\times P$ so $f_\ast:T(M\times P)\to T(M\times P)$ $\endgroup$ – JonHerman Sep 24 '14 at 2:02
  • $\begingroup$ My mistake on the *. And you are right about what I meant. $\endgroup$ – AndresB Sep 24 '14 at 2:44
  • $\begingroup$ In general, the pushforward cannot be decomposed, but it can if the function has the form $f(m,p) = (g(m), h(p))$ for functions $g$, $h$. (The converse is probably also true, but I don't see how to prove it immediately). If you'd like an examples of an $f$ not of that form, just consider $f:S^1\times S^1\rightarrow S^1\times S^1$ with $f(z,w) = (z,zw)$ (where I'm thinking of $S^1$ as the unit complex numbers). $\endgroup$ – Jason DeVito Sep 24 '14 at 2:46
  • $\begingroup$ JasonDevito. Thanks for replying. I´m not seeing at the moment whyt he push forward of that function can not be decomposed. Still, I find it interesting because I'm actually interested in manifold that is a product of two Lie groups, Namely $SU(n)\times SU(n)$. If $(g,g)\in SU(n)\times SU(n)$, then I would like to define $L_{(1,h)}(g,g)\text{´}=(g,hg)$ and then to find it push forward. $\endgroup$ – AndresB Sep 24 '14 at 3:00
  • $\begingroup$ Suppose $f:M\times P\to M\times P$. Let $f_1=\pi_1\circ f$ and $f_2=\pi_2\circ f$ where $\pi_1:M\times P\to M$ and $\pi_2:M\times P\to P$ are the projection maps. For arbitrary $(p,q)$ you can say locally that $f_\ast=f_{1,\ast}+f_{2,\ast}$ which is something in $T_pM$ plus something in $T_qM$. Is this what you're after? This is only a local result, so it's not really an answer to your question $\endgroup$ – JonHerman Sep 24 '14 at 3:03

Not entirely sure if this is what you're asking for or not, but let's give this a go.

Observe that at any point $(p,q)\in P\times Q$ we can write $$T_{(p,q)}P\times Q\simeq T_pP\oplus T_qQ,$$ where $T_qQ$ is imbedded in $T_{(p,q)}P\times Q$ as the tangent space of the submanifold $\{p\}\times Q$, and similarly for $T_pP$.

Then if $\phi :P\times Q\to P\times Q$ is a smooth map, and $\phi(p,q)=(p',q')$, its differential gives a map $$\phi_* : T_pP\oplus T_qQ \to T_{p'}P\oplus T_{q'}Q.$$ Therefore we can write $\phi_*$ as a matrix $$\phi_* = \begin{pmatrix} A & B \\ C & D \end{pmatrix},$$ where $$A : T_pP\to T_{p'}P,$$ $$B:T_qQ\to T_{p'}P,$$ $$C:T_pP\to T_{q'}Q,$$ $$D:T_qQ\to T_{q'}Q.$$

You can check that if we unravel the definitions, we have $A$ is the differential of the map $P\to P'$ defined by $p\mapsto \pi_1(\phi(p,q))$, i.e., holding $q$ constant. Similarly, $B$ is the differential of the map $Q\to P$ given by $q\mapsto \pi_1(\phi(p,q))$. We get similar results for $C$ and $D$.

Let's not focus too much on the details, since I think this will become more clear if we apply it to your particular case of interest.

Let $G$ and $H$ be Lie groups, $\alpha : G\to H$ a Lie group homomorphism. Define $\phi : G\times H \to G\times H$ by $(g,h)\mapsto (g,\alpha(g)h)$. We want to compute its differential.

$A$ is the differential of the map $g\mapsto g$, since we're holding $h$ constant and projecting onto the first factor. Thus $A=1$ is the identity map.

$B$ is the differential of the map $h\mapsto g$, holding $g$ constant. Thus $B=0$.

$C$ is the differential of the map $g\mapsto \alpha(g)h$, holding $h$ constant, so $C=r_{h*}\alpha_*$.

Finally $D$ is the differential of the map $h\mapsto \alpha(g)h$, holding $g$ constant, so $D=\ell_{\alpha(g)*}$.

Thus we have that the differential of $(g,h)\mapsto (g,\alpha(g)h)$ is given at a point $(g,h)$ by $$\begin{pmatrix} 1 & 0 \\ r_{h*}\alpha_* & \ell_{\alpha(g)*} \end{pmatrix}.$$

In the particular case where $G\subseteq H$, so $\alpha$ is just the inclusion, we have that the differential is $$\begin{pmatrix} 1 & 0 \\ r_{h*} & \ell_{g*} \end{pmatrix}.$$

If $X\in \mathfrak{g}\subseteq \mathfrak{h}$, $Y\in\mathfrak{h}$, then in this last case, we have $$\phi_*(X,Y) = \begin{pmatrix} 1 & 0 \\ r_{h*} & \ell_{g*} \end{pmatrix}\begin{pmatrix} X \\ Y\end{pmatrix} = \begin{pmatrix} X \\ Xh + gY \end{pmatrix}$$

If we're identifying the tangent spaces via left multiplication, then $gY_h\in T_{gh}H$ corresponds to the same element of $\mathfrak{h}$ as $Y_h$, and $X_gh=gX_eh = ghh^{-1}X_eh = gh\operatorname{Ad}_{h^{-1}}X_e$.

Thus we end up with $\phi_*(X,Y) = (X,\operatorname{Ad}_{h^{-1}}X+Y)$.


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