0
$\begingroup$

This is probably a super naïve question, but it seems like the only reason $f(x)=\sqrt{x}$ isn't continuous on $\mathbb{R}$ is because it pops over $\mathbb{C}$ when $x<0$. So, seeing as $\mathbb{R}\subset\mathbb{C}$, is $f$ continuous there?

I feel like there's probably a gigantic part of math that's dedicated to the study of this sort of thing, but I don't have sufficient background to know what that is. Any elucidation would be awesome!

$\endgroup$
  • 2
    $\begingroup$ To define $\sqrt{z}$ for $z$ a complex number, the"most natural" definition would be to set, for $z = re^{i 2\theta},$ $\sqrt{z} = \sqrt{r} e^{i \theta},$ where $\sqrt{r}$ is the square root of non-negative real numbers. This "definition" is bogus for $\theta$ is not uniquely determined and this is where you choose a "branch" for the argument; this however, makes it impossible to define $\sqrt{z}$ continuous in the entirety of the complex plane. $\endgroup$ – Will M. Apr 15 at 18:21
  • 1
    $\begingroup$ What I am expressing is that $\sqrt{z}$ is not defined until you set up what branch you choose and then, it is defined. As far as I am concerned, $\sqrt{z}$ will be continuous on the real axis for some branches. $\endgroup$ – Will M. Apr 15 at 18:24
  • 1
    $\begingroup$ You can study the square root function on Riemann surfaces as multivalued function, so it‘s definitely not continuous! $\endgroup$ – Fakemistake Apr 15 at 18:26
  • $\begingroup$ Interesting! So then for which branches would $\sqrt{z}$ would be continuous on the real axis? Is an identification of such branches ever useful? $\endgroup$ – yungblud Apr 15 at 18:31
  • $\begingroup$ On any simply connected subset of $\mathbb C \setminus \{0\}$, there is a continuous branch of $\sqrt{z}$. More generally, there is such a continuous branch in any domain where $0, \infty$ are in the same component of the complement. $\endgroup$ – GEdgar Apr 15 at 19:49
2
$\begingroup$

Since the $f(z)=\sqrt{z}$ is not single-valued (check!), it is not even well-defined in $\mathbb{C}$. There are two ways to get around this:

  • Riemann surfaces: We basically construct a space in which our function is single-valued, usually resulting in some cool picture (there are nice visuals on Youtube). The Riemann surface varies depending on which function we are working with, e.g. $\sqrt{z}$ or $w^2=z(z-1)(z-\lambda)$. Formally, a Riemann surface is a one-dimensional complex manifold (in particular they are all algebraic curves!) and there is a lot of beautiful theory about them, but let's not get ahead of ourselves.
  • Branch cuts: We "cut" a part of the complex plane out of our domain of definition. The popular one for $\sqrt{z}$ is cutting out the negative real axis, so we have a plane with an infinite "slit." This also occurs with functions such as $\log{z}$ (fun fact: this cut is so popular it is called the principal branch cut!). The only issue with doing this is that when we integrate along some contour $\gamma$, we must make sure that it does not hit the cut. For example, if we cut out the negative real axis we can't integrate along a circle - the way we get around this is by using what's called the "keyhole" contour.
$\endgroup$
  • 1
    $\begingroup$ Worth to add: if you define $\sqrt{z}$ on its Riemann surface, it's continuous. If you define it on $\mathbb{C}$ with a cut, then it's not, and the cut is where it's discontinuous. $\endgroup$ – Adam Latosiński Apr 15 at 20:09
1
$\begingroup$

Here's the rundown of the various cases we might consider when asking "is the square root function continuous?":

$(1)$ We can define $\sqrt{\cdot}:\mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ in the standard way, and this is continuous.

$(2)$ We cannot continuously define $\sqrt{\cdot}: \mathbb{R} \to \mathbb{R}$; we cannot define this at all, in fact, as there is no sense in which negative numbers have real square roots.

$(3)$ We can define $\sqrt{\cdot}: \mathbb{C} \to \mathbb{C}$, but we cannot do so continuously essentially because attempting to do so inevitably leads to a contradiction when one travels around a circle in the complex plane (other answers and the comments have more details).

$(4)$ We can continuously define $\sqrt{\cdot}:\mathbb{R} \to \mathbb{C}$ in a few (four) different ways, the most "natural" one being $\sqrt{-r}=i\sqrt{r}$ for $r>0$ with the standard definition of $\sqrt{\cdot}$ on $\mathbb{R}_{\geq 0}$-- check the definition of continuity holds!

So, there are several natural ways we can try to interpret the square root function, and whether or not it is continuous depends on this interpretation. Less natural to the casual observer but the most complete way to view the picture is through the corresponding Riemann surface, which captures the full structure of $\sqrt{\cdot}$ on $\mathbb{C}$ while maintaining continuity by actually considering a larger domain than $\mathbb{C}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.