Yesterday I came across this webpage, which describes a recent (successful) attempt to visualize isometric embeddings of flat tori in 3D Euclidean space. The webpage and associated paper discuss the difficulty in creating this visualization, which was why it hadn't been done before. Are there other problems or objects in mathematics that are similarly hard to visualize?
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$\begingroup$ Groups are hard to visualize precisely, though there are vague visualizations that are useful for proof intuition. $\endgroup$– QuditApr 19, 2015 at 20:50
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$\begingroup$ This question seems too broad as written; when even embeddings of surfaces in three dimensions are hard to visualise, there seems to be little hope for the vast majority of mathematics! $\endgroup$– user64687Apr 19, 2015 at 23:06
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$\begingroup$ Regarding flat rectangular tori in $\mathbf{R}^{3}$...it's trivial to roll a paper rectangle into a cylinder, press it flat, then roll the resulting "tubular band" into a cylinder to create a flat torus. Naturally the result isn't immersed, but the flat, toroidal geometry is perfectly apparent. :) $\endgroup$– Andrew D. HwangApr 19, 2015 at 23:51
1 Answer
Yes, there are plenty. This is true even if we restrict ourselves to geometrical objects. On one hand, visualizing geometrical objects in $4$ or more dimensions can be (an usually is) very challenging, even for simple ones like the tesseract (alias $4$-hypercube):
On the other hand, we can only really depict geometry over the real numbers. For example, how do you visualize the locus $x + 2 y = 0$ with $x,y \in \Bbb{C}$? (hint: a complex plane!) What about the locus $x^2 + y^2 = r$?
Even worse, getting a decent idea of what a scheme (a kind of geometrical object) looks like can be quite hard. A classical example (an one of my favourites) is the picture of $\text{Spec}(\Bbb{Z}[X])$ from Mumford's red book
of which you can find a nice explanation in this old blog post.
Note that there is quite a difference between these examples and the one you cited, though, in that these are (in a sense) even "harder" to visualize.
What I mean is that these objects cannot be faithfully represented in the "usual" $2$ or $3$ dimensional geometry, and all we can do is associate some images to them from which we can glean some (hopefully a lot) useful information).
On the other hand, what I understand from the webpage you cited is that the method used by Nash and Kuiper to prove the existence of isometric embeddings of flat tori in $3$ dimensional space was ineffective. This is a bit like being able to prove that a certain quantity is bounded, but being unable to tell exactly what the bound is. Sure, you cannot provide an exact picture following the method described in that page because you would have to depict a limit shape, but this is a finer issue: you can still get a really good idea of what that shape is and, given enough computational power, you can approximate it arbitrarily better.
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$\begingroup$ I don't see the point of the word "locus". They're just subsets. Why use it? $\endgroup$ Apr 19, 2015 at 23:24
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1$\begingroup$ @goblin $x+2y=0$ is not a set, it's an equation. for example, $2x+4y=0$ is a different equation, yet... $\endgroup$– AlbertApr 19, 2015 at 23:29
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$\begingroup$ @Glougloubarbaki, A.P. is using "locus" to refer to the subset defined by the equation, not the equation itself. $\endgroup$ Apr 19, 2015 at 23:30
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$\begingroup$ yes, so would you rather say "the subset of those $(x,y)$ satisfying this equation" or "the locus of this equation" ? I guess it's just a question of vocabulary $\endgroup$– AlbertApr 19, 2015 at 23:34
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$\begingroup$ @goblin I am referring to the what is called the zero locus of a polynomial $f(x,y)$ (or, equivalently, of an equation $f(x,y) = 0$). Formally, this is the pair $(f, V(f))$ where $V(f)$ is the subset of $(x,y) \in \Bbb{C}^2$ such that $f(x,y) = 0$. Practically, this term is used for the set $V(f)$, with the proviso of distinguishing between different equations if necessary. More precisely, in (classical) algebraic geometry the object of study are the zero loci, not the underlying algebraic sets. $\endgroup$– A.P.Apr 19, 2015 at 23:35