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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Groups are hard to visualize precisely, though there are vague visualizations that are useful for proof intuition. – Qudit Apr 19 '15 at 20:50
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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! – Apr 19 '15 at 23:06
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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. :) – Andrew D. Hwang Apr 19 '15 at 23:51
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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
![picture of Spec(Z[X])](../../images/b061cba550f675403b7f133ad0a67370.webp)
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.
A.P.
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I don't see the point of the word "locus". They're just subsets. Why use it? – goblin GONE Apr 19 '15 at 23:24
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1@goblin $x+2y=0$ is not a set, it's an equation. for example, $2x+4y=0$ is a different equation, yet... – Albert Apr 19 '15 at 23:29
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@Glougloubarbaki, A.P. is using "locus" to refer to the subset defined by the equation, not the equation itself. – goblin GONE Apr 19 '15 at 23:30
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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 – Albert Apr 19 '15 at 23:34
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@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. – A.P. Apr 19 '15 at 23:35
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@A.P. I didn't know that. But I don't really see the point of this, either - the pair $(f,V(f))$ contains the same amount of information as $f$. Why not just deal with $f$? Or just deal with $f$ modulo scaling by elements of the base ring that have two-sided multiplicative inverses. – goblin GONE Apr 20 '15 at 00:03
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@goblin Because $f$ is an algebraic object, while $V(f)$ is a geometric one. A parabola isn't just a quadratic polynomial, after all... Usually the main interest is the geometric object, and we use an algebraic description of it to gain more insight. It's like studying a smooth manifold and using its sheaf of smooth functions to understand more of it -- e.g. to define/describe its tangent bundle. – A.P. Apr 20 '15 at 08:13
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@A.P., here's what I suspect is really going on. Let $f$ denote a polynomial in $k$-many variables, with coefficients in a ring $R$. Then--when defined correctly--$V_f$ is the function (functor?) that assigns to every $R$-algebra $A$ the set $V_f(A)$, defined as ${a \in A^k \mid f(a) = 0}$. That way, $V_f$ will potentially remember details about $f$ that $V_f(R)$ doesn't. What do you think; is this a good way of looking at it? – goblin GONE Apr 20 '15 at 11:11
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@goblin Yes, it is: you just discovered the functor of points of the scheme $\text{Spec}(R[x_1,\dotsc,x_k] / \langle f \rangle)$! (also, see this example) – A.P. Apr 20 '15 at 11:36
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