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 $D_x g$?

Thanks for reading!

  • 1
    $\begingroup$ 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? $\endgroup$ – Thomas Rot Aug 7 at 15:11
  • $\begingroup$ @ThomasRot that seems like a very reasonable suggestion. Is it possible to use this to obtain something coordinate free? $\endgroup$ – Ben Aug 7 at 15:36
  • $\begingroup$ @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? $\endgroup$ – Ben Aug 7 at 15:38
  • $\begingroup$ Spensers argument is what I had in mind. This is reasonably coordinate free $\endgroup$ – Thomas Rot Aug 7 at 15:38
  • $\begingroup$ I hadn't seen it until now! Please ignore that $\endgroup$ – 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 $$D_x(fg)=D_x(m\circ(f\times g))=D_{(f(x),g(x))}m\circ(D_xf\times D_xg)=f(x)D_xg+g(x)D_xf.$$

  • $\begingroup$ Marvellous! I believe the canonical identification (plus inexperience) was what was tripping me up there. That does seem jolly reasonable. $\endgroup$ – Ben Aug 7 at 15:45
  • $\begingroup$ @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 $C^\infty(V)\to\mathbb{R}:f\mapsto\left.\frac{d}{dt}\right|_{t=0}f(v+tw)$. $\endgroup$ – Spenser Aug 7 at 15:47
  • $\begingroup$ Of course! That makes a lot of sense $\endgroup$ – Ben Aug 7 at 16:33

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