# Smooth maps from given manifold into the complex plane: is there a “product rule”?

Suppose that we have two smooth functions from some real manifold $$X$$ into the complex plane $$\mathbb C$$:

$$f,g: X \to \mathbb C.$$

I'm vaguely wondering, if we treat $$\mathbb C$$ as a plane equipped with a flat metric, if there's a relationship between (complex) multiplication and the Frechet derivative map, i.e. a kind of "product rule" which relates $$h(x):= f(x)g(x)$$ to $$f$$ and $$g$$.

My super naive thinking would be that something like $$(\ast)$$ holds:

$$\phantom{(\ast)} \qquad D_x h = g(x)D_x f + f(x) D_x g \qquad (\ast)$$

(where $$D_x f : T_x X \to T_{f(x)} \mathbb C$$ denotes the Frechet derivative of $$f$$, for example),

but this doesn't even make sense, as $$D_x f$$, $$D_x h$$ and $$D_x g$$ can all land in different tangent spaces. This leads to my question.

Question: Is there any meaningful way to relate the derivative map of $$h$$ in terms of $$f$$ and $$g$$, given that there one can identify tangent spaces of $$\mathbb C$$ using isometries?

Failing that, is it possible to at least estimate the operator norm of $$D_x h$$ from that of $$D_x f$$ and $$D_x g$$?

• What is the derivative of the map $\mathbb C\times \mathbb C\rightarrow C$ that takes the product? How can you use this and the chain rule? – Thomas Rot Aug 7 at 15:11
• @ThomasRot that seems like a very reasonable suggestion. Is it possible to use this to obtain something coordinate free? – Ben Aug 7 at 15:36
• @rschwieb I'm not assuming anything like that because I don't know what it means. Would it be at all helpful to assume that they are? – Ben Aug 7 at 15:38
• Spensers argument is what I had in mind. This is reasonably coordinate free – Thomas Rot Aug 7 at 15:38
• I hadn't seen it until now! Please ignore that – Ben Aug 7 at 15:40

Since $$\mathbb{C}$$ is a vector space, its tangent spaces can be canonically identified with itself. Hence, for each $$x\in X$$, we may view $$D_xf$$ as a map $$D_xf:T_xX\to\mathbb{C}.$$ Then, $$D_x(fg)=f(x)D_xg+g(x)D_xf.$$ To prove this formula, note that $$fg=m\circ(f\times g)$$ where $$m:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$$ is multiplication and $$f\times g:X\to\mathbb{C}\times\mathbb{C}$$ is the map $$(f\times g)(x)=(f(x),g(x))$$. Then, $$D_x(f\times g)=D_xf\times D_xg:T_xX\to\mathbb{C}\times\mathbb{C}$$. Moreover, after identifying $$T_z\mathbb{C}$$ with $$\mathbb{C}$$ for $$z\in\mathbb{C}$$ we have that for all $$(z_1,z_2)\in\mathbb{C}\times\mathbb{C}$$, $$D_{(z_1,z_2)}m:\mathbb{C}\times\mathbb{C}\to\mathbb{C},\quad (a,b)\mapsto z_1b+z_2a$$ since $$(D_{(z_1,z_2)}m)(a,b)=\left.\frac{d}{dt}\right|_{t=0}(z_1+at)(z_2+bt)=z_1b+z_2a.$$ Combining this with the chain rule, we get $$D_x(fg)=D_x(m\circ(f\times g))=D_{(f(x),g(x))}m\circ(D_xf\times D_xg)=f(x)D_xg+g(x)D_xf.$$
• @Ben My pleasure. By the way, if $V$ is a vector space and $v\in V$, the canonical identification between $V$ and $T_vV$ is the linear isomorphism $V\to T_vV$ which sends a vector $w$ to the derivation $C^\infty(V)\to\mathbb{R}:f\mapsto\left.\frac{d}{dt}\right|_{t=0}f(v+tw)$. – Spenser Aug 7 at 15:47