# Push Forward on product manifold.

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 $$TP$$

• 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 $f_\ast:T(M\times P)\to T(M\times P)$ – JonHerman Sep 24 '14 at 2:02
• My mistake on the *. And you are right about what I meant. – AndresB Sep 24 '14 at 2:44
• 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). – Jason DeVito Sep 24 '14 at 2:46
• 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. – AndresB Sep 24 '14 at 3:00
• 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 – 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 $$\phi_*(X,Y) = (X,\operatorname{Ad}_{h^{-1}}X+Y)$$.