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I'm having trouble understanding connectedness from a group theoretic perspective.

Let $G$ be the symplectic group of dimension 4 over a field $K$,

$$G = \operatorname{Sp}_4(K) = \left\{ A \in \operatorname{GL}_4(K) : A^T J A = J \right\} \text{ where } J = \left(\begin{smallmatrix}.&.&.&1\\.&.&1&.\\.&-1&.&.\\-1&.&.&.\end{smallmatrix}\right)$$

and let $C$ be the centralizer of a specific unipotent element $t$,

$$C=C_G(t) \text{ where } t = \left(\begin{smallmatrix}1&1&.&.\\.&1&.&.\\.&.&1&1\\.&.&.&1\end{smallmatrix}\right)$$

The exercise asks one to, Show that $t$ does not lie in the connected component of the identity when the characteristic of $K$ is 2. I think K is algebraically closed, though this is perhaps not specified here (and is specified in a nearby exercise).

I calculate the centralizers to be:

$$C_{\operatorname{GL}_4(K)}(t) = \left\{ \left(\begin{smallmatrix}a&b&c&d\\.&a&.&c\\e&f&g&h\\.&e&.&g\end{smallmatrix}\right) : a,b,c,d,e,f,g,h \in K, ag-ec \neq 0 \right\} \cong \operatorname{GL}_2\left(K[dx]/{(dx)}^2\right)$$ $$C_{\operatorname{Sp}_4(K)}(t) = \left\{ \left(\begin{smallmatrix}a&b&c&d\\.&a&.&c\\e&f&g&h\\.&e&.&g\end{smallmatrix}\right) : a,b,c,d,e,f,g,h \in K, ag-ec = 1, ah+bg=cf+de \right\}$$

I am clueless how to find their connected components.

What are the connected components of $C_{\operatorname{GL}_4(K)}(t)$ and $C_{\operatorname{Sp}_4(K)}(t)$?

Especially describe the exceptional behavior in characteristic 2.

Does the connectedness have anything to do with them being matrices?

I would prefer some group theoretic way to find the components, but I worry that the components have nothing to do with the matrices, and depend only on the equations $ag-ec=1$ and $ah+bg=cf+de$, regardless of where these variables are in the matrix.

If they don't have anything to do with the group structure, then why would I care if it is connected?

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I am even more clueless: I don't know what connected components even mean in the absence of topology. (And I don't know where a topology on K is going to come from)... A common way to prove that elements x and y lie in different components is to exhibit a continuous real valued function f which is never zero and satisfies $f(x)<0<f(y)$. –  user31373 Jul 17 '12 at 4:53
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@LeonidKovalev: the topology doesn't seem to come coordinate by coordinate (from K), but rather fairly directly on the 8-tuples (or 16-tuples, I'm not sure) from the Zariski topology (which ensures that points are closed and polynomials are continuous). The centralizers are both closed, as they are the solutions of polynomial equations. Using some multi-variate polynomial factoring ideas, one can decide connectedness, but it seems unrelated to the group structure. This is troublesome due to the large number of group theory theorems that definitely require connectedness. –  Jack Schmidt Jul 17 '12 at 15:29
    
@JackSchmidt: It seems like there must be a mistake in your calculation of $C_{Sp_4(K)}(t)$ because the element $t$ is not in the set on the RHS. –  Michael Joyce Sep 9 '12 at 23:49
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@MichaelJoyce: There is no mistake in that part. If the characteristic of K is 2, then t is in the RHS, and if the characteristic of K is not 2, then t is not in Sp4(K) so is not in the RHS. –  Jack Schmidt Sep 9 '12 at 23:53
    
Ahh, I see now. Thanks. –  Michael Joyce Sep 10 '12 at 0:00
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2 Answers

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+500

I wrote up my findings in this short note. In brief: the exercise is incorrect. Both centralizers are connected. I confirmed with the authors that the intention of the exercise was an exhibition of Springer's 1966 result that a regular element does not lie in the connected component of its centralizer in a few cases, including symplectic groups in characteristic 2. Unfortunately the element given is not regular. The short note contains an example regular element, a calculation of its centralizer, and the decomposition into its connected components.

To find the connected components one has to factor the defining equations. For $C_{GL}$ this is easy: they are irreducible, and the entire centralizer is its own irreducible component. For $C_{Sp}$ this is a bit more challenging as the two polynomials together still define an irreducible system of polynomial equations. The exercise was intended to be easy: for the regular element, the centralizer factors into two obviously irreducible systems.

Since writing up the solution, I have had help proving the centralizer is connected using more elementary methods, mostly using pre-1920s mathematics. However, a few of the details are still too complicated to justify giving here (where I believe one would like a clear and simple explanation for LVK, and a group theoretic explanation for me).

As far as the importance of connectedness in group theory, this remains a mystery. Certainly no-one who helped me over the course of several weeks used the group structure to solve it, though I believe it could be possible to do so.

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I think the claim in the OP is mistaken, because of the following : define for $t \in K$,

$$ p(t)= \left(\begin{smallmatrix}1&t&.0&0.\\0.&1&0.&0.\\0.&0.&1&-t\\.0&0.&0.&1\end{smallmatrix}\right) $$

Then $p$ is polynomial and hence continuous for the Zariski topology. As the image of a connected set by a continuous map stays connected, we deduce that $p(K)$ is connected. So $p(O)=I$ and $p(1)=t$ are in the same connected component, (they are in fact "arcwise connected") contrary to the original claim.

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