# One-step problems in geometry

I'm collecting advanced exercises in geometry. Ideally, each exercise should be solved by one trick and this trick should be useful elsewhere (say it gives an essential idea in some theory).

If you have a problem like this please post it here.

Remarks:

• I'm collecting such problems for many years. The current collection is here, it is about 80 problems.

• At the moment, I have just few problems in topology and in geometric group theory and just one in algebraic geometry.

• Thank you all for nice problems --- I gave bounty to one, but would give it to 4 problems if I could :). Some of them are in the collection by now --- please post more...

-
If anyone wonders why I deleted my previous comment, I looked more closely at Anton's file of examples and realized his exercises were at a higher level than I had thought. –  David Speyer Dec 8 '09 at 23:31
Just a comment and vote up to say nice collection of problems. I have to restrain myself to not go attempt to solve them all and instead study for my topology qual. –  B. Bischof Dec 9 '09 at 4:14

## 16 Answers

Here is a problem that I learned from W. Thurston. I do not remember whose problem it was originally. Possibly Conway?

Suppose that you have a finite collection of round circles in round $S^3$, not necessarily all of the same radius, such that each pair is linked exactly once. (In particular, no two intersect.) Prove that there is an isotopy in the space of such collections of circles so that afterwards, they are all great circles.

-
Nice problem, I solved it :) But I can not see where else such idea can be useful... –  Anton Petrunin Dec 14 '09 at 4:24
One proof uses intersection of planes with a 4-ball. Arguably this leads to ideas such as hyperbolic geometry in n+1 dimensions to study spheres in n dimensions, as well as the 4-ball definition of linking in 3 dimensions. –  Greg Kuperberg Dec 14 '09 at 4:40
Accepted (my solutions was not that clever). –  Anton Petrunin Dec 14 '09 at 5:00
Hey Greg, I think I remember hearing this problem mentioned by Oleg Viro when he gave a talk at Davis when I was there. Could this be where Thurston got it? –  Ian Agol Dec 17 '09 at 19:08
Maybe. My recollection is that he cited a discussion between him and Conway. We'd have to ask Bill to find out, I guess. –  Greg Kuperberg Dec 17 '09 at 19:29

Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:

There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?

To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)

P.S. When I was asked the question, I failed miserably!

-
Yes, it is a nice problem --- thanks:). I heard nearly the same question from David Berg (UIUC). He was interested in geodesics in rigions of Euclidean space cutted by a graph of Lipschitz function. If I remember right, he has a paper on this subject. –  Anton Petrunin Dec 9 '09 at 18:16
Shh! Say "geodesic" and you'll give the answer away! –  Joel Fine Dec 9 '09 at 18:35
Here is the ref: Berg, I. D. An estimate on the total curvature of a geodesic in Euclidean 3-space-with-boundary. Geom. Dedicata 13 (1982), no. 1, 1--6. –  Anton Petrunin Dec 11 '09 at 3:01
@Darsh, don't worry, everyone else Frederic asked on the same train journey also failed first time. –  Joel Fine Dec 13 '09 at 23:42
A lasso's loop is meant to contract when pulled tight. –  horse with no name Sep 25 '11 at 19:19

Anton asks for more problems in topology and algebraic geometry. One issue is that the concept of a "trick" is treated differently in these two areas than in differential geometry. In topology, not quite as many ideas are called "tricks"; they are sometimes named after people and co-opted as material, e.g., the Alexander trick and the Whitney trick. In algebraic geometry, tricks are sometimes regarded as suspect; they are sometimes taken as a reason to reorganize definitions to either again co-opt the trick or avoid it outright.

Still, a problem based on the Alexander trick could be at a good level for this problem list.

Problem: Prove that space of tame knots, meaning piecewise-linear embeddings $f:S^1 \to \mathbb{R}^3$, is connected in the $C^0$ topology on functions $f$.

-
I will take this one, thank you :) –  Anton Petrunin Dec 18 '09 at 14:30

Does Steve Fisk's "trick" in solving the "Art Gallery Problem" of Victor Klee qualify?

Show that for a simple plane polygon with n sides, floor function (n/3) vertex guards are sometimes necessary and always sufficient to "see" all of the interior points of the polygon.

-
Sounds great, I'll try to solve it first... –  Anton Petrunin Dec 9 '09 at 4:28

Since Gjergji brings up isoperimetric inequalities, there is a lot of attention in combinatorial geometry devoted to combinatorial versions. For instance, if $f$ is a Boolean function $f$ on $n$ bits, define its "instability" to be the number of ways that $f(\vec{x}) \ne f(\vec{y})$ when $\vec{x}$ and $\vec{y}$ differ in one bit. If half of the values of $f$ are 0 and half are 1, then the theorem is that the most stable choice of $f$ is a function $f(\vec{x}) = x_k$ that only depends on one bit.

On the theme of one-step proofs in geometry that depend on another theorem, there is a one-step proof of this fact using the standard spherical isoperimetric inequality.

-
I did not solve it --- any hints? –  Anton Petrunin Dec 18 '09 at 14:32
The domain of $f$ is the set of vertices of an $n$-cube. Make a correspondence between these vertices and orthant subsets of the sphere in $n-1$ dimensions. –  Greg Kuperberg Dec 18 '09 at 15:28

Since you asked for a problem in algebraic geometry, here is a popular result whose proof in modern terms is very short. It could be called a one-step problem:

Harnack's inequality on curves: Prove that a smooth algebraic curve of degree $d$ in $\mathbb{R}P^2$ consists of at most $(d^2-3d+4)/2$ circles. (1,1,2,4,7,11,...)

-
I like the problem, usually I sign it by author or give a ref where a nice solution can be found (or at least a hint). Any suggestions? –  Anton Petrunin Dec 19 '09 at 13:38
I do not have a good reference other than that Wikipedia has a page called the Harnack curve theorem. I learned the problem in graduate school from a paper, but I do not remember which paper. As for a hint: A smooth complex curve of degree $d$ has genus $g = (d^2-3d+2)/2$. This looks sophisticated to prove, but it is not very hard. It may look sophisticated to use that to prove Harnack's bound $g+1$, but that's not very hard either. –  Greg Kuperberg Dec 19 '09 at 16:03

Suppose that we have two simple closed curves in $R^3$ which are linked. And suppose that the distance between these curves is $1$. Prove that the length of each curve is at least $2\pi$.

This problem has an interesting history. It was published in the book by W. Hayman, Unsolved problems in Function theory, where it was attributed to F. Gehring. I solved it in 1977, jointly with Oleg Vinkovski, prepared a paper and gave a seminar talk. After the talk, I was approached by an undergraduate student, who proposed a ridiculously simple solution. Just two lines, using nothing. So I did not submit my paper. Later I've seen several published solutions, but none of them was so simple.

EDIT. Here is this proof (due to Igor Syutrik). Fix a point $M$ on $A$. Then one can find another point $M'$ on $A$ such that the interval $[M,M']$ intersects $B$. Indeed, otherwise we can deform $A$ to $M$ moving straight along these intervals $[M,M']$ and deformation will not cross $B$. Let $O$ be a point on $[M,M']$ that belongs to $B$. Let $A'$ be the central projection of $A$ from $O$ onto the unit sphere around $O$. Then $A'$ passes through two diametrically opposite points of the sphere and thus its length is at least $2\pi$.

-
There is a 4-sentence solution attributed to Marvin Ortel written up in the first few lines of the paper "Criticality for the Gehring link problem" by Cantarella et al arxiv.org/abs/math.DG/0402212 Is it by chance the one you heard? –  j.c. Nov 11 '12 at 16:13
Yes, it is a good one, thank you. Is the proof linked by jc is the same as the proof of the graduate student you mentioned? –  Anton Petrunin Nov 11 '12 at 18:17
Yes, this is the same proof:-) But the student was UNDERgraduate:-) I add, that Gehring, when stating the problem, added that he could prove that $length(A)\geq c$, where $c$ is an absolute constant. No one EXPECTED that the solution could be THAT simple. –  Alexandre Eremenko Nov 11 '12 at 19:57
The situation reminds me the famous problem of J-J. Sylvester: "Can we have a configuration of finitely many lines in the (real) projective plane such that all intersections are triple." (There is such a configuration in the complex projective plane $C^2$.) In the beginning of XX century this was a famous unsolved problem, sometimes listed next to the "4-colors problem". In 1944 it was solved, and the solution is such that it could be found by a clever high school kid. Also about 2 lines of text, using nothing. –  Alexandre Eremenko Nov 11 '12 at 20:08
@AlexandreEremenko, Google found this math.purdue.edu/~eremenko/dvi/gehring.pdf for me. Are you planning to submit it to arXiv? (Otherwise it is hard to use as a reference.) –  Anton Petrunin Sep 26 '13 at 16:37

I believe this is a problem given in J. Hirsch's Differential Topology. This may be much simpler than the ones posted here already. But for what it's worth, here it is.

Show that given a collection of spheres the product manifold embeds into an Euclidean space of one dimension higher, viz., for instance $S^2 \times S^3$ embeds in $\mathbb{R}^6$.

-

Problem: Consider two closed smooth strictly convex planar curves, one inside another. Show that there is a chord of the outer curve, which is tangent to the inner curve and divided by the point of tangency into equal parts.

-
Not bad --- I solved it :). BTW do you know that a similar problem is open: find a point on outer curve which has two tangent segments from this point to the inner curve has equal size (at least if outer curve is not convex). –  Anton Petrunin Dec 29 '09 at 4:33
I did not know that problem and could not solve it (even if outer curve is convex). I can comment it: Each point x of inner curve defines "left" and "right" segment of a chord tangent at x and corresponding values l(x) and r(x). Consider images L and R of functions l and r correspondingly. If L \subset R or R \subset L then for any diffeomorphism g of the inner curve there exists x such that L(g(x))=R(x). It seems that, in general, it is not true that L \subset R either R \subset L. –  Petya Jan 1 '10 at 18:01
I can prove your problem for a case when l (or r) is a constant function. Another variant (generalization) of the initial problem, unknown to me, is the following. Consider (locally) affine coordinate x on the inner curve, which is an angle of a tangent line with a fixed line. This coordinate is defined uniquely up to a sum with constant. Then, for any angle a an equation l(x)=r(x+a) is solvable. –  Petya Jan 1 '10 at 18:14
Petya, this statement which you can do is everything known about the problem (if I remember right my conversation with Sergei Tabachnikov). –  Anton Petrunin Jan 30 '10 at 19:38

Measure concentration for the sphere $S^n$, is a nice problem if the use of a theorem (Brun-Minkowski inequality) counts as a trick.

Problem: Let $\mu$ be the normalized measure in the unit sphere $S^n$. Let $X$ be a measurable subset such that $\mu(X)=\frac12$, and $X_{\delta}$ be the set of all $x\in S^n$ for which there is $x'\in X$ such that $||x-x'||=\delta$. Prove that $\mu(X_{\delta})\geq 1-2e^{-\frac{n\delta^2}{4}}$. (So increasing $X$ by just a little gives almost the entire sphere.)

On a similar line stands the "Classical Isoperimetric Inequality". (Among all bodies of the same volume, the ball has the least surface area.)

Another fact that comes to mind (with the trick being a compactness argument) is Bunt-Motzkin theorem:

Problem: Show that a simple polytope $P\subset \mathbb{R}^n$ is convex iff for every point $p$ not in $P$, there exists a unique point in $P$ that's closest to $p$.

-
(1) I do not know a proof of first problem which use Brun--Minkowski :( Can you give a hint? (2) Second problems looks nice --- I have to think a bit :)... Thank you –  Anton Petrunin Dec 14 '09 at 4:36
A possible hint: Let Y be the complement of $X_{\delta}$. Now form $X'$ as all $ax$ with $x\in X,a\in [0,1]$, similarly for $Y'$. What can be said about $X',Y'$ and $\frac{X'+Y'}{2}$ as subsets of $\mathbb{R}^{n+1}$? –  Gjergji Zaimi Dec 17 '09 at 7:36

Greg Kuperberg's comment on Alexander tricks reminded me of a nice one due to Tom Goodwillie.

Let $K_n$ be the space of $C^k$-smooth embeddings of $\mathbb R$ into $\mathbb R^n$ in the $C^k$-topology $k>1$, where the embeddings are required to be 'long' in the sense that $f(t)=(t,0,\cdots,0)$ for $t \notin [-1,1]$. Let $Imm_n$ be the corresponding space of long immersions $\mathbb R \to \mathbb R^n$.

Then the inclusion map $K_n \to Imm_n$ is null-homotopic. It's a one-line proof provided you know the Smale-Hirsch theorem. Or if you want to remove Smale-Hirsch, replace $Imm_n$ by $\Omega S^{n-1}$ and let the map $K_n \to \Omega S^{n-1}$ be the normalized velocity vector.

Similarly, there's a nice one-line proof that the inclusion map $K_n \to K_{n+1}$ is null-homotopic. The original idea is ancient but this formulation (as far as I know) is due to me. You don't need any theorems for this, it's a construction for which you can write down the null-homotopy using simple functions.

-

I think the question Is it possible to capture a sphere in a knot? is an excellent one-step problem.

-
A variation of this problem is already there (1.11. Stable net). The original problem requires two steps... –  Anton Petrunin Dec 29 '09 at 22:15

This is admittedly not very hard even without getting the trick, but it's super-easy with the trick. Hopefully it isn't too easy to be interesting (or even amusing):

Problem. Let $v\_1,\ldots,v\_n$ be vectors in $\mathbb{R}^m$, and let $V$ be the $m\times n$ matrix whose columns are $v\_1,\ldots,v\_n$. Show that the $n$-dimensional volume of the $n$-dimensional parallelepiped in $\mathbb{R}^m$ determined by $v\_1,\ldots,v\_n$ is $\sqrt{\det(V^TV)}$.

We should probably take as given that the determinant of a square matrix is the signed volume of the parallelepiped determined by its columns (or by its rows); I would consider justifying that to be a separate problem (for which I don't know any simple tricks).

-

I like the following problem. It has a very short solution based on a (very) useful trick (idea).

We fix three pairwise tangent (at distinct points) spheres $A_1$, $A_2$ and $A_3$ in 3-space. Let us construct a sequence of spheres: first sphere $B_1$ is a (generic) sphere which is tangent to each $A_i$. $B_2$ is a sphere which is tangent to $B_1$ and to each $A_i$ (in fact there are two such spheres, we chose one of them as a $B_2$). $B_3\ne B_1$ is a sphere which is tangent to $B_2$ and to each $A_i$ (such a sphere is unique), etc: $B_{n+1}\ne B_{n-1}$ is a sphere which is tangent to $B_n$ and to each $A_i$.

The problem is to show that $B_7=B_1$.

-
Nice problem, but it is more for school students; the target is graduate students... –  Anton Petrunin Mar 15 '10 at 18:46
I have some statistic... I think it is a hard problem even for professionals. You have good graduate students! (Vladimir Igorevich could say it is a problem for a kindergarten.) –  Petya Mar 16 '10 at 2:33

Every finite point set has Delauney triangulation (circumsphere of any simplex doesn't contain points from this set).

-
Yes, I deside to include one problem on Delauney triangulation (it is 6.6 "inscribed triangulation"). –  Anton Petrunin Mar 16 '10 at 1:22
I mean another proof. There is very nice proof through stereographic projection on sphere –  akopyan Mar 16 '10 at 2:42
Nice proof, I did not know it before :) –  Anton Petrunin Mar 16 '10 at 18:06
-