Gromov: why we don’t explain things

This common and unfortunate fact of the lack of adequate presentation of basic ideas and motivations of almost any mathematical theory is probably due to the binary nature of mathematical perception.

Either you have no inkling of an idea, or, once you have understood it, the very idea appears so embarrassingly obvious that you feel reluctant to say it aloud…

(Gromov, “Stability and Pinching”)

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9 thoughts on “Gromov: why we don’t explain things”

1. Mark Meckes says:

2. mozibur ullah says:

Very true.

3. plm says:

I remember reading and thinking about this quote so many times, I am not sure where it comes from now. Perhaps from Pier’s edited book “Development of mathematics: 1950-2000″, Rémi Langevin interviews Gromov, or an interview by Marcel Berger.

This is a tendency I personally always struggle against, that of considering complicated facts trivial. I have felt this keenly in a course in analytic number theory I have taken last semester, here in Bordeaux. There was just a massive amount of deep ideas, and they are often presented as utterly trivial.

As example, after a talk by Minhyong Kim at Michael Taylor’s conference in Bordeaux this summer, where he mentioned that there is no integral point on the projective line minus {0,1,\infty}, I remember asking the teacher of the above course why, because I obviously thought of 2 as an integral point. Annoyed he replied approximately “Take the reduction mod 2, it gives a point on the projective line over $\mathbb F_2$ but there is none if you remove {0,1,\infty}.”.

The origin of such attitude is probably in great part the huge pressure required to learn and do mathematics productively, pushing the limits of human intellect, of motivation, to specialize in seemingly useless, abstract, but extremely complex reasoning.

I think there is no easy solution, my attitude is to go slowly and insist on insights, as I feel some of my models as mathematicians did. Grothendieck for instance, insisted on finding the right framework for solving a problem, and on not having a despiseful attitude, rather on taking a childlike attitude, at several levels.

In doing this I have troubles with exams (especially written) which demand speed, and with some of my teachers. For sure, part of the troubles are not due to imitating great mathematicians like Grothendieck. :) But I think researchers should make an effort to understand nonspecialists, should be more tolerant.

PS: The key idea why “1″ (and -1) is still an integral point, in the situation above, but not “2″ nor other integers is that we can define a morphism from $(\text{Spec }\mathbb Z[T])\setminus V(\{0\})=\text{Spec }\mathbb Z[T,T^{-1}]$ to $\mathbb Z$ evaluating polynomials at 1 (and -1), but not at 2, nor any other integer $n$, because the ideal generated in $\mathbb Z$ will be $(n)\neq\mathbb Z$, so the image of the morphism $\text{Spec }\mathbb Z\rightarrow\text{Spec }\mathbb Z[T]$ will intersect $\text{Spec }\mathbb Z[T,T^{-1}]=\left\{(T),(T,p): p\text{ prime}\right\}$, at $(T,p)$, $p|n$. Of course a solution that is certainly well known to experts is to consider “S-integral points”, points which are defined except at a finite number of places (morphisms from $\text{Spec }\mathbb Z[p_1^{-1},p_2^{-1},...,p_s^{-1}]$), then we do recover all equations with integral coefficients as points on the integral affine line minus the origin. And if we remove 2 or finitely many S-integral points we still have infinitely many. I think this is the notion of integral point that should be given in books. I’d be glad to hear opinions because I am not very confident about this, understanding it really requires concentration and computations, care.

(Hoping the latex comes out ok…)

4. plm says:

Addendum, it’s misleading when I say “the ideal generated in $\mathbb Z$ will be $(n)\neq\mathbb Z$“. Morphisms are determined setting $T=n$ (this is the universal property of polynomial rings), the intersection with $V(\{0\})$ is the ideal generated by the constant term of $(T-n)$ and $T$, ${(T,n)}$, so I meant $\mathbb Z$ considered as $\mathbb Z[T]/(T)$, not the image of $ev_n:\mathbb Z[T]\rightarrow\mathbb Z$ which is Z. I have been confused with this, I hope I got it right.

5. BCnrd says:

Dear plm: There’s a more “functorial” (and I think cleaner) way to understand the situation with your question about the integral points, without needing to haul out the ring homomorphisms explicitly or make it look as if $\mathbf{F}_2$ plays a special role. Inside the projective line over any base at all, the complement of the section $\infty$ is the affine line, and inside the affine line the complement of the section $0$ is the multiplicative group scheme ${\rm{GL}}_1$, and finally inside the multiplicative group scheme the complement of the section $1$ is the functor of units $u$ such that $u-1$ is also a unit (which is just another way of saying that $u \not\equiv 1 \bmod \mathfrak{p}$ for all prime ideals $\mathfrak{p}$ of $R$). So for any ring $R$ at all, the $R$-valued points of the complement of $\{0, 1, \infty\}$ is the set of units $u \in R^{\times}$ such that $u-1 \in R^{\times}$. (One can say a similar thing for maps from any manifold into the complement of $\{0,1\infty\}$ in the real projective line.) There are no such units in $\mathbf{Z}$. Note that the same reasoning applies with $\mathbf{Z}$ replaced by any imaginary quadratic field, even those for which 2 is inert (so the argument with $\mathbf{F}_2$ is no longer applicable). QED

6. BCnrd says:

Oops, near the end of my comment above I meant to say “replaced by the ring of integers of any imaginary quadratic field”.

7. plm says:

Dear BCnrd,

I wish you were my teacher. :)

We have Liu Qing as mathoverflow person here but he does not teach -and actually I think I might annoy him as a student.

Also, I correct that the conference was in honor of Sir Martin J. Taylor, not Michael.

Thanks alot.

PS: just add “latex” after the opening dollar sign to typeset latex formulas in wordpress.

8. NDE says:

Well, there is *one* imaginary quadratic field for which the complement of {0,1,∞} has a couple of integral points…

9. NDE says:

Also there’s nothing wrong with the argument using reduction mod 2 (except for the part about being annoyed). An S-integral point of the complement of {0,1,∞} is one that whose reduction mod p falls outside {0,1,∞} for all primes p outside S. So if there is such a point then S must contain all p with 2-element residue field.