Connectedness of centralizer exercise 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?
 A: 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.
A: 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.
