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This is something that I find is always a bit vague in differential geometry and would be very glad if someone could give me a definite rule.

Here is a prototype example of what I want to compute. Let $A,B,C,D$ some smooth manifolds and $$ f:A\to B\\ g:A\to C\\ h:B\times C\to D $$ now we can construct a map $F:A\to D$ in the following way: $$ A\ni p\mapsto F(p)=h(f(A),g(A))\in D $$ My question is how to express the differential $D_pF$ in terms of $Df$, $Dg$ and $Dh$? I would like to see something like the chain rule $D_p(f\circ g)=D_g(p)f.D_pg$ but don't know how to do this for multi-variable functions like $h$ in the above example.

Edit

Thanks to AlexR and martini this question has been answered: $$ D_pF.X_p = D_{f(p),g(p)}h.(D_pf.X_p,D_pg.X_p) $$ There is a follow-up question "Differential on a product space as sum of differentials" on how to rewrite this expression a sum.

Community
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Stan
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    Note that $D_{f(p), g(p)} h$ lives on $T_{B\times C}$. Try to find $D_p \tilde f$ for $$\tilde f: A\to B\times C, a \mapsto (f(a), g(a))$$ Then $F = h\circ \tilde f$. – AlexR Jul 17 '14 at 09:57

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$F$ can be written as $$ F = h \circ (f \times g) \circ \Delta_A $$ where $f \times g \colon A \times A \to B \times C$ is the map $(f\times g)(a,a') = \bigl(f(a), g(a')\bigr)$, with differential $$ D_{(a,a')}(f\times g) = D_af \times D_{a'}g $$ (using the identification $T(B \times C) \cong TB \times TC$ of the tangent bundles, induced by the differentials of the projections. $\Delta_A$ is the diagonal embedding $A \ni a \mapsto (a,a) \in A^2$, which has $\Delta_{TA}$ as its differential (that can again be seen using the coordinate projections). Now, by the chain rule $$ DF = Dh \circ (Df \times Dg) \circ \Delta_{TA} $$ That is for $(p,X) \in TA$, we have $$ D_pF(X) = Dh_{f(p),g(p)}\bigl(D_pf(X),D_pg(X)\bigr)$$

martini
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  • Thank, it was very useful to know that $\Delta_A$ has to be used, as well as that $D\Delta_A=\Delta_{TA}$. – Stan Jul 17 '14 at 12:06
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    Just for other people who look for the proof of the latter: $D_a\Delta_A(X_a)=\left.d/dt\right|{t=0}\Delta_A(\gamma(t))=\left.d/dt\right|{t=0}(\gamma(t),\gamma(t))=(X_a,X_a)=(\Delta_{TA}X)_a$ where $\gamma(t)$ a path such that $\gamma(0)=a$ and $\dot\gamma(0)=X_a$. – Stan Jul 17 '14 at 12:06