diffgeom,
@diffgeom@mathstodon.xyz avatar

Three-sheet Monty, branched over the origin: The complex cubing map (top to bottom) and multi-valued cube root (bottom to top) branched along the negative real axis.

Complex analysis books generally describe the Riemann surface of the cube root as something like "three copies of the slit complex plane, with the lower edge of each cut joined cyclically to the upper edge of the next cut." This description is correct, but (for me, at least) hides the simple global picture: The Riemann surface of the multi-valued cube root function is itself a complex plane.

The discontinuity of the principal cube root across the branch cut is depicted geometrically in the top plane by the jump in position of the larger dot. The continuity of the multi-valued function is similarly depicted as a rotating equilateral triangle of cube roots.

Comparable pictures hold for square roots, fourth roots, etc.

An animation loop depicting the complex cubing function and cube root multi-function against a black background. The plane at bottom is a gold rectangle with a blue Cartesian grid. The plane at top is a curvilinear dodecagon, the image of the Cartesian grid under the cube root. The principal cube root has the colors of the bottom, blue on gold. The other two branches, multiplied by non-trivial cube roots of unity, are gold on blue.

narain,
@narain@mathstodon.xyz avatar

@diffgeom Very nice picture, and it makes a lot of sense!

Your use of the phrase "continuity of the multi-valued function" piqued my interest. Is there a general theory of continuity of multi-valued functions? I've had reason to wonder about this from time to time, but never enough to really look into it.

diffgeom,
@diffgeom@mathstodon.xyz avatar

@narain One natural approach comes from quotienting the domain of a mapping into level sets.

Formally, if (f:X \to Y) is a continuous surjection, we define an equivalence relation on (X) by (x \sim x') if and only if (f(x) = f(x')), whose equivalence classes are the levels of (f). The induced mapping (\bar{f}:X/R \to Y) is continuous and bijective, its inverse "is" the multi-valued inverse of (f), and we can ask if the inverse is continuous, which happens if (f) is an open mapping.

Here with (X = Y) the complex numbers and (f) the cubing map, the quotient is particularly nice; my claim about the multi-valued cube root can be interpreted in these terms.

Generally, non-constant holomorphic maps in one variable are open, and non-constant entire functions are either surjective or omit precisely one value by the big Picard theorem, so this framework works well for Riemann surfaces of inverse functions. :)

narain,
@narain@mathstodon.xyz avatar

@diffgeom I wish I knew enough topology and complex analysis to find this intuitive 😅

I was thinking along these lines. One wants to be able to say, for example, that the map from a matrix to its eigenvalues is continuous. How would one formalize that? In fact this is a many-to-many map, so I'm not sure it's possible to perform a useful quotient here.

diffgeom,
@diffgeom@mathstodon.xyz avatar

@narain The answer of Georges Elencwajg to this question at math.stackexchange is worth a read if you haven't seen it. :)
https://math.stackexchange.com/questions/63196/continuity-of-the-roots-of-a-polynomial-in-terms-of-its-coefficients

Implicit in Henning Makholm's comment just below the question is a wrinkle if we're working over the reals, which are not algebraically closed: The zero matrix, for example, is "close to" matrices with pairs of "nearly zero" real roots and to matrices with no real roots. Still, in the sense the real square root multi-function is continuous at (0), it looks to me that we get continuity of eigenvalues in the entries for real matrices by restricting the eigenvalue behavior of complex matrices (and ignoring non-real eigenvalues).

If you're asking about operators on infinite-dimensional spaces, however, we need a functional analyst....

narain,
@narain@mathstodon.xyz avatar

@diffgeom Interesting, I didn't get a notification for this post. It doesn't show up in my replies/mentions tab either, even though I'm clearly tagged. @christianp, is this a bug in @Mastodon ?

christianp,
@christianp@mathstodon.xyz avatar

@narain @diffgeom weird! There's nothing in the job queue, so it's not just waiting to happen.
Not sure what went wrong. I got a notification about your mention of me, for what it's worth.

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