Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 167 / 2

An element $g\in G(k)$ is semisimple when its image in a faithful finite-dimensional representation is diagonalizable, and unipotent when that image has every eigenvalue equal to one. These conditions are independent of the faithful representation. The Jordan decomposition in an affine algebraic group is the unique factorizationwith $g_s$ semisimple and $g_u$ unipotent.

In $G={\mathrm{GL}}_2$, matrices with two distinct eigenvalues form a dense open subset and are semisimple, so semisimple elements are dense. Their full locus is not open: a scalar matrix is semisimple, but every neighbourhood of it contains a nontrivial Jordan block. The group also has nonidentity unipotent matrices.

For a contrasting example take $G={\mathbb{G}}_a\times {\mathbb{G}}_m$. Its semisimple locus is $\{0\}\times {\mathbb{G}}_m$, which is closed and not dense, while $(a,1)$ for $a\ne 0$ is unipotent and $(0,t)$ for $t\ne 1$ is semisimple.

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The derived subgroup of an affine algebraic group $[G,G]$ is the closed subgroup generated by the commutators $xy{x}^{-1}{y}^{-1}$. If $G$ is connected, the image of every finite product of commutator maps is connected and contains the identity. The closures of these images form an increasing sequence; once their dimensions stabilize, the stable member is closed under products and inverses and equals $[G,G]$. Hence $[G,G]$ is connected.

Now suppose the connected group $G$ is solvable. The Lie-Kolchin theorem conjugates a faithful representation of $G$ into the upper triangular matrices. Every commutator then has all diagonal entries equal to one, so every element of $[G,G]$ is unipotent. Moreover $[G,G]$ lies in the upper unitriangular group, whose superdiagonal filtration is a central series. It is therefore a nilpotent group.

A diagonalizable algebraic group is a closed subgroup of a product of copies of ${\mathbb{G}}_m$. A unipotent algebraic group has only unipotent elements, while a semisimple algebraic group here means one all of whose elements are semisimple. A reductive algebraic group is smooth, connected, affine, and has trivial connected normal unipotent radical.

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The Kolchin theorem conjugates a faithful representation of a unipotent algebraic group $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication givesSince $U_n^{(n)}=1$, this filtration is a finite central series, so $U_n$ and every subgroup of it are nilpotent groups. Hence $U$ is nilpotent.

The converse fails: ${\mathbb{G}}_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.

Solved by gpt-5.6-sol high.