Differential on a product space as sum of differentials

Question

I would like to derive a rule for differentiating maps of the form $h(f(\bullet),g(\bullet))$ on smooth manifolds which is equivalent to partial differentials of multi-variable functions on $\mathbb R$.

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: $$ A\ni p\mapsto F(p)=h(f(A),g(A))\in D $$ From the question "Differential of a multi-variable map" we know that the differential of $F$ is given by $$ D_pF.X_p = D_{f(p),g(p)}h.(D_pf.X_p,D_pg.X_p) $$ Question: is it legitimate to rewrite it as follows: $$ D_pF.X_p = D_{f(p)}h(\bullet,g(p)).D_pf.X_p + D_{g(p)}h(f(p),\bullet).D_pg.X_p\,? $$ And how to come up with this form without using the intuition and analogy to regular functions as in the following example: let $f(x)$, $g(x)$ and $h(y_1,y_2)$ $\mathbb R$-valued functions on $\mathbb R$, then (with some obvious abuse of notation): $$ \partial_xh(f(x),g(x)) = \partial_{y_1}h(f(x),g(x))\partial_xf(x) + \partial_{y_2}h(f(x),g(x))\partial_xg(x) $$

Answer 1

Levap gave a great answer and I would like to summarize and put it in my own words.

We want to prove the following claim:

Claim: let $B$, $C$, and $D$ smooth manifolds and $$ h:B\times C\to D $$ a smooth map. Then, for any $(X,Y)\in\mathfrak X(B\times C)$ we have $$ \boxed{D_{b,c}h.(X,Y) = D_bh(\bullet,c).X + D_ch(b,\bullet).Y} $$

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Proof: $Dh$ is a linear map and $(X,Y)_p = (X,0)_p + (0,Y)_p$ for all $p\in B\times C$, therefore $$ D_{b,c}h.(X,Y) = D_{b,c}h.(X,0) + D_{b,c}h.(0,Y) $$ Now take smooth paths $\gamma_X$ and $\gamma_Y$ in $B$ and $C$ respectively, such that $$ \gamma_X(0) = b\qquad\dot\gamma_X(0) = X_b\\ \gamma_Y(0) = c\qquad\dot\gamma_Y(0) = Y_c\\ $$ then $\gamma_X\times\{c\}$ and $\{b\}\times\gamma_Y$ are paths in $B\times C$ through $(b,c)$ with velocities $(X,0)_{b,c}$ and $(0,Y)_{b,c}$. Then we can write the right-hand-side as: $$ \left.\frac{d}{dt}\right|_{t=0}h(\gamma_X,c) + \left.\frac{d}{dt}\right|_{t=0}h(b,\gamma_Y) $$ but this is nothing else than the right-hand-side of the claim.$\quad\square$

Answer 2

This is indeed legitimate. To justify it rigorously without getting too boggled down with notation, I suggest that you do first the case where $A = (-\varepsilon, \varepsilon)$, $D = B \times C$, $h = \mathrm{id}$, $p = 0$ and $X_p = \left. 1 \cdot \frac{d}{dt} \right|_{t=0}$. That is, handle first the case of curves.

Rename $f = \beta$ and $g = \gamma$. Then, what you wrote reads as follows:

$$ \left. \frac{d}{dt} F(t) \right|_{t=0} = \left. \frac{d}{dt} h(\beta(t),\gamma(t)) \right|_{t=0} = \left. \frac{d}{dt} (\beta(t),\gamma(t)) \right|_{t=0} = \left. \frac{d}{dt} (\beta(t),\gamma(0)) \right|_{t=0} + \left. \frac{d}{dt} (\beta(0), \gamma(t)) \right|_{t=0}. $$

You can prove it by writing everything in local coordinates (you won't even need the chain rule). This is also the thing behind the isomorphism $T_{(b,c)} B \times C \cong T_bB \oplus T_cC$. How does the isomorphism work? Given a tangent vector $X_b \in T_bB$ represented by a curve $\beta(t)$ and a tangent vector $Y_c \in T_cC$ represented by a curve $\gamma(t)$, the corresponding tangent vector in $T_{(b,c)}B \times C$ is given by $\left. \frac{d}{dt} (\beta(t),\gamma(t)) \right|_{t=0}$. However, since in $T_bB \oplus T_cC$ we have $(X_b,Y_c) = (X_b,0_b) + (0_c,Y_c)$, the isomorphism should respect this and so we must also have $$ (X_b,Y_c) = \left. \frac{d}{dt} (\beta(t),c) \right|_{t=0} + \left. \frac{d}{dt} (b,\gamma(t)) \right|_{t=0}. $$

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For the general case, by using the definition of the differential as something that acts on curves, we can reduce it to the previous case. Let $\alpha \colon I \rightarrow A$ be a smooth curve such that $\alpha(0) = p$ and $\dot{\alpha}(0) = X_p$. Then $$ D_pF(X_p) = \left. \frac{d}{dt} F(\alpha(t)) \right|_{t=0} = \left. \frac{d}{dt} h(f(\alpha(t)),g(\alpha(t))) \right|_{t=0}. $$ We want to show need to show that this is equal to $$ \left. \frac{d}{dt} h(f(\alpha(t)),g(p)) \right|_{t=0} + \left. \frac{d}{dt} h(f(p),g(\alpha(t))) \right|_{t=0} $$ but this follows immediately from the case of curves discussed above and the regular chain rule.