In studying deformation theory of Galois representations, I've come surely to an error, relating Schlessinger's criterion.

Let's fix a representation $\bar{\rho}$ of a group $G$ and let $D_{\bar{\rho}}$ be its deformation functor from the category $\hat{\mathcal{C}}$ of complete noetherian local $W(\mathbb{F})$-algebras (where $W(\mathbb{F})$ is the ring of the Witt vectors over a finite field $\mathbb{F}$) with residue field $\mathbb{F}$. Let $A'$, $A''$ and $A$ be artinian objects of $\hat{\mathcal{C}}$, and let be given two homomorphisms of algebras $u' : A' \rightarrow A$ and $u'': A'' \rightarrow A$. One of the conditions of Schlessinger's criterion, namely that the natural map $$D_{\bar{\rho}}(A'\times_A A'') \rightarrow D_{\bar{\rho}}(A') \times_{D_{\bar{\rho}}(A)} D_{\bar{\rho}}(A'')$$ should be surjective when $u'$ is a small surjection, seems to me always true, regardless the surjectivity of $u'$. Take in fact $([\rho'], [\rho'']) \in D_{\bar{\rho}}(A') \times_{D_{\bar{\rho}}(A)} D_{\bar{\rho}}(A'')$ and let $\rho: G \rightarrow \operatorname{GL}_n(A'\times_A A'')$ be defined as $\rho(\sigma) = (\rho'(\sigma), \rho''(\sigma))$, using the identification $$\operatorname{GL}_n(A'\times_A A'') \simeq \operatorname{GL}_n(A')\times_{\operatorname{GL}_n(A)}\operatorname{GL}_n(A'').$$ By the property of this last identification, we see that by conjugating one of $\rho'$ or $\rho''$, we conjugate also $\rho$. Thus this yields a well defined $[\rho] \in D_{\bar{\rho}}(A'\times_A A'')$ and seems just the one we wanted.

Since this must clearly be wrong, where is my mistake?

Sorry if this question reveals my big miscomprehension! Thank you in advance.


I think the "mistake" is in the definition of the equivalence relation that defines a representation. Indeed, an $A$-valued representation (for some $A\in\hat{C}$) is a conjugacy class of homomorphisms $$ \rho:G\to \operatorname{GL}_n(A) $$ reducing to $\bar{\rho}$. This makes the equality $$ D_\bar{\rho}(A'\times_AA'')=D_\bar{\rho}(A')\times_{D_\bar{\rho}(A)}D_{\bar{\rho}}(A'') $$ false, in general, because the sets $D_\bar{\rho}(\cdot)$ are smaller than you would expect (and this can make the fiber product behave oddly). The point is that two homomorphisms $\rho'\colon G\to\operatorname{GL}_n(A')$ and $\rho''\colon G\to\operatorname{GL}_n(A'')$ might reduce to different morphisms in $\operatorname{GL}_n(A)$ (so, the pair does not define an element in the left-hand side) but these reductions might be conjugate in $\mathrm{GL}_n(A)$ thus belonging to the right-hand side. Surjectivity of $u\colon A'\to A$ helps you out of this trouble.

A word of warning: actually, what I wrote above is slightly incorrect. It would be correct if we were considering the functor assigning to each $A$ the set of $A$-valued representations of $G$ with values in $\operatorname{GL}_n(A)$; but the functor $D_\bar{\rho}$ one normally considers while deforming Galois representations is a bit different, namely you are allowed to conjugate only by matrices in $$ \operatorname{Ker}\Big(\operatorname{GL}_n(A)\to\mathrm{GL}_n(\mathbb{F})\Big). $$ In other words, if $\rho,\rho'$ verify $M\rho(g)M^{-1}=\rho'(g)$ for some $M\in\operatorname{GL}_n(A)$ and all $g\in G$, then the representations $\rho,\rho'$ are isomorphic but they define the same deformation only if $M\equiv \mathrm{id}.\pmod{m_A}$ where $m_A$ is the maximal ideal of $A$. In particular, the set $D_\bar{\rho}(A)$ is even bigger than before.

You find a detailed proof that surjectivity of $u'$ implies what you want in the right language of deformations (instead of mere representations) for instance in Section 3.1 of Tilouine's Deformations of Galois Representations and Hecke Algebras, Mehta Res. Institute (1995); on page 391 of his original Deformations paper in Galois groups over $\mathbb{Q}$ Mazur gives an extremely succinted discussion - actually, he just says "it is easy", so it might be of small use.

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  • $\begingroup$ Thank you! I just missed the fact that my $(\rho'(\sigma), \rho''(\sigma))$ is NOT in general an element of $GL_n(A'\times_A A'')$, because of the conjugacy in $GL_n(A)$. Thank you a lot! $\endgroup$ – Simone Melchiorre Chiarello Feb 16 '14 at 8:36
  • $\begingroup$ But sorry, is it then always true if $u'$ is surjective? If the images of $\rho'$ and $\rho''$ in $A$ are conjugated by an element $S \in GL_n(A)$, then by surjectivity there exists an element $S' \in GL_n(A')$ that is sent by $u'$ in $S$. Then the couple $(S'\rho'(S')^{-1}, \rho'')$ is indeed in $GL_n(A'\times_A A'')$ by construction. $\endgroup$ – Simone Melchiorre Chiarello Feb 16 '14 at 14:37
  • $\begingroup$ My be my answer was not very clear. What I want to stress is that your functor is different from $A\mapsto\mathrm{GL}_n(A)$ and that this causes troubles when performing the fiber product. Now I do not understand exactly your question (or, at least, it seems that you found the answer yourself...) $\endgroup$ – Filippo Alberto Edoardo Feb 16 '14 at 16:01
  • $\begingroup$ In my last comment I have proved that if $u'$ is surjective, then the map $D_{\bar{\rho}}(A' \times_A A'') \rightarrow D_{\bar{\rho}}(A')\times_{D_{\bar{\rho}}(A)} D_{\bar{\rho}}(A'')$ is surjective. This must be wrong: where is it wrong? One more question: I don't see why injectivity can also fail. A trivial verification makes me believe that it is always injective, and obviously it is not true... I'm really sorry for this but I'm new to the subject. $\endgroup$ – Simone Melchiorre Chiarello Feb 16 '14 at 16:07
  • $\begingroup$ I see: well, you are right in saying that if $u'$ is surjective then we are done (so, condition H2 in Schlessinger's paper is vacuous for us): it is false for very general functors like the ones in Schlessinger's papers but true for deformation ones and your argument is right. For injectivity, this has really to do with the fact that in deformation business you only allow conjugation by matrices which are $1\pmod{m_A}$. If you check the references in my answer you find the full argument (and understand where injectivity can fail, Tilouine's proof is quite transparent). $\endgroup$ – Filippo Alberto Edoardo Feb 16 '14 at 16:25

i think you miss the condition 4 of the Schlessinger's criterion :

(*) $Dρ¯(A′×_{A}A′′)→Dρ¯(A′)×_{Dρ¯(A)}Dρ¯(A′′)$

(*) is a bijection whenever $A′ = A”$ is a small extension !

see : http://www.math.jussieu.fr/~harris/SatoTate/notes/Schlessinger.pdf

Here we start with $ρ$ and $ρ′$ over the fiber product such that $ρ_{A}$ and $ρ_{B}$ are conjugate to $ρ^′_{A}$ and $ρ^′_{B}$ respectively.

Choose conjugators $M_{A}$ and $M_{B}$. Note that if $˜M_{A}$ and $˜M_{B}$ were equal, we could lift them to $Gl_{n}(A ×_{C} B)$ and we would be done. This is true in general with no hypotheses on A,B, and C.

see : http://www.math.ias.edu/~bwlevin/Schlessinger.pdf

so only when (*) is a bijection, we can lift the morphism to its representation !

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