I am reading from the book "Topics in Galois Theory" by Serre.

I came across the word "Rigidity". I am not able to understand this concept.

If I am not wrong, This term was first used by Thompson, He used this concept("Rigidity Method") to show that monster group can be realized as the Galois group over $\mathbb{Q}$.

I don't have any intuition for this concept. I tried reading from the Wikipedia page on "Rigidity", but I got more confused. It says

The above statement does not define a mathematical property. Instead, it describes in what sense the adjective rigid is typically used in mathematics, by mathematicians.

What is the idea ? One more question, What does the adjective "rigid" means in context of Galois theory.

  • 1
    $\begingroup$ There is a precise notion of rigidity in group theory which Thompson used to prove that many specific groups, including the monster group, are Galois groups of Galois extensions over $\mathbb{Q}$. See the section on rigidity in the Wikipedia page on the inverse Galois problem for details and references: en.wikipedia.org/wiki/Inverse_Galois_problem $\endgroup$ Nov 1, 2016 at 23:31
  • $\begingroup$ There's also Mostow's rigidity theorem, very different context though. $\endgroup$ Nov 2, 2016 at 15:54

1 Answer 1


Consider a (ramified) Galois covering of the Riemann sphere $f\colon X\to B=\mathbf P^1(\mathbf C)$, of group $G$. Let $b_1,\dots,b_r\in B $ be the (distinct) ramification points.

Fix a base point $o\in B\setminus\{b_1,\dots,b_r\}$ and for all $i$, fix an injective $C^1$-path $\gamma_i$ from $o$ to $b_i$ that they do not intersect outside of $o$ (in particular, they avoid the other ramification points). Define an element $\sigma_i$ in $\pi_1(B,o)$ by the following recipe: start from $o$, go along the chosen path until very close to $b_i$, make one turn around $b_i$ in the positive direction, and go back to $o$ along the chosen path. If the cyclic ordering of the points $b_1,\dots,b_r$ is chosen so as to coincide with the angles the paths leaving $o$, then the elements $\sigma_1,\dots,\sigma_r$ generate the group $\pi_1(B,o)$, and there is the single relation $\sigma_1\cdots \sigma_r=e$. (This reflects in particular the fact that the Riemann sphere, and even the complex plane, is simply connected.)

The group $\pi_1(B,o)$ acts naturally on $X$ (on the right) via an anti-homomorphism $\pi_1(B,o)\to G$.

Fix a point $x\in X$ above $o$. The path $\gamma_i$ lifts to a path from $x$ to some point $a_i$ in $X$. The subgroup $G_i$ of $G$ that fixes $a_i$ is cyclic of order $e_i$ (the ramification index at $i$), generated by the action $[\sigma_i]$ of the loop $\sigma_i$.

In particular, $[\sigma_1],\dots,[\sigma_r]$ generate $G$.

The elements $[\sigma_1],...,[\sigma_r]$ depend on the choice of the initial paths $\gamma_i$. Changing those paths turns them into conjugate elements $h_i [\sigma_i] h_i^{-1}$. In particular, their conjugacy classes $C_i$ are well-defined, and $C_1,\dots,C_r$ generate $G$.

In general, if $G$ is a finite group, one says that a sequence $(C_1,\dots,C_r)$ of conjugacy classes is rigid if:

(i). For each $i$, there exists $g_i\in C_i$ such that $g_1\dots g_r=e$ and $(g_1,\dots,g_r)$ generate $G$;

(ii). For any two families $(g_1,\dots,g_r)$ and $(g'_1,\dots,g'_r)$ as in (i) are conjugate by an element $h\in G$: $g'_i=h g_i h^{-1}$ for all $i$.

The interest in the concept — take the time to study Serre's book if you need so — is that the Inverse Galois Problem is “easy” to solve in these cases. The arguments is basically as follows. First of all, one constructs a covering of the Riemann sphere satisfying these geometric properties, where the points $b_i$ are integers, say. By Riemann's existence theorem, this corresponds to a finite field extension $\mathbf C(T)\subset F$, Galois of group $G$, with inertia subgroups corresponding to the given conjugacy classes. Now, consider an arbitrary automorphism of $\mathbf C$; thanks to the rigidity property, it naturally extends to an automorphism of $F$. All these automorphisms allow to construct a canonical subfield $E$ of $F$ such that there is a finite field extension $\mathbf Q(T)\subset E$, that gives $F$ after extending the scalars from $\mathbf Q$ to $\mathbf C$. (I am slightly cheating here, because some cyclotomic extension may enter the picture. Let's pretend there is none, which happens when the classes $C_i$ are “$\mathbf Q$-rational”.)

Let $e\in E$ be a primitive element which is, say, integral over $\mathbf Q[T]$. Its minimal polynomial $P$ belongs to $\mathbf Q[T,X]$, and is irreducible viewed as a polynomial in $\mathbf C[T,X]$. Now apply Hilbert's irreducibility theorem: for many elements $u\in\mathbf Q$ (“most of them”, in some sense), the polynomial $P(u,X)$ is irreducible in $\mathbf Q[X]$ and generates a Galois extension of $\mathbf Q$ of group $G$. The Inverse Galois problem is solved in this case.


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