# Diagonalizable linear algebraic group is isomorphic to $(\mathbb{C}^*)^r\times A$, for some finite abelian group $A$

I have three questions about algebraic groups.

Let $D$ be a linear algebraic group. Then the following are equivalent:

1. $D$ is diagonalizable.

2. $\mathop{Hom}(D,\mathbb{C}^*)$ is finitely generated with an isomorphism of coordinate rings $\mathbb{C}[D]\cong \mathbb{C}[\mathop{Hom}(D,\mathbb{C}^*)]$.

3. Every rational representation of $D$ is isomorphic to a direct sum of $1$-dimensional representations.

4. $D$ is isomorphic to $(\mathbb{C}^*)^r\times A$, for some finite abelian group $A$.

I'm fine with #1-3, but I had to pause after reading #4 because I am now wondering why and how does this torsion part arise (it may be because I do not know any algebraic groups other than the ones that can (naturally) be embedded into $GL(n,\mathbb{C})$).

Example: Take $D$ to be a diagonal group with character group $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$. Then the coordinate ring of $D$ is $\mathbb{C}[x,x^{-1},y]/\langle y^2-1\rangle$ with $D\cong \mathbb{C}^*\times \mu_2$.

$\mathbf{Question \;1}$: Is there a way to write $D\cong \mathbb{C}^*\times \mu_2$ in a single matrix form, i.e., embed $D$ into $GL(n,\mathbb{F})$? Or must it be separately embedded into a product of $GL_n$'s like $GL(1,\mathbb{C})\times GL(1,\mathbb{F}_2)$?

Suppose our above $D\cong \mathbb{C}^*\times \mu_2$ acts on $\mathbb{C}^3$ by $$(x,y).(a,b,c)=(xa,x^{-1}b,yc).$$

$\mathbf{Question\; 2}$: Then do we have $$(x,y)^2.(a,b,c)=(x^2a,x^{-2}b,c) \mbox{ with } \mathbb{C}[a,b,c]^D=\mathbb{C}[ab,c^2]?$$

$\bf{Question \; 3}$: Can you give me an example of an algebraic group which cannot be embedded into a product of $GL_n$'s?

Thank you.

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I don't understand 4. "Isomorphic" as what? The right hand side is an abelian group, the left hand side is a matrix; matrices are not abelian groups, though they can be used in several different ways to define abelian groups. What kind of isomorphism are we talking about? –  Arturo Magidin Jul 11 '12 at 3:53
If $D$ is a diagonal matrix, then it is isomorphic to $(\mathbb{C}^*)^r$ as varieties and as groups, right? I was reading a few lecture notes and they are referring to books by Humphreys, Springer, and Borel. –  math-visitor Jul 11 '12 at 4:02
You mean that the \emph{set} of diagonal matrices is isomorphic to $(\mathbb C^*)^r$, not a single matrix. –  Grumpy Parsnip Jul 11 '12 at 4:04
Correct. If $D$ is the set of all diagonalizable matrices in $GL_n$, then we can construct a group homomorphism which is an isomorphism which is also a map of varieties. –  math-visitor Jul 11 '12 at 4:04
Sorry Arturo. I misunderstood a few times. Yes, an example would be and I meant the set of diagonal matrices in $GL(r,\mathbb{C})$ is isomorphic to $(\mathbb{C}^*)^r$. –  math-visitor Jul 11 '12 at 4:09

## 2 Answers

Some comments on question 2. Your answer is correct. It is a question in classical invariant theory. $C^*\times \mu_2 \cong GL(1)\times O(1)$, act on $\mathbb{C}[\mathbb{C} \oplus \mathbb{C}^*]\otimes \mathbb{C}[\mathbb{C}]$ one each factor respectively. Classical invariant theory tell you the $GL(n)$-invariant in $\mathbb{C}[\mathbb{C}^n\oplus (\mathbb{C}^n)^*]$-the first factor is generated by $<a,b>$ (the natural paring) and the $O(n)$ invariant in $\mathbb{C}[\mathbb{C}^n]$-the second factor is generated by $c^2$. But it is not obvious there is no further relations between the quadratic invariants. Fortunately, your case $n=1$, is in stable range, so no further relations.

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Thank you Hoxide! Quick question: how does one get $\mathbb{C}[\mathbb{C}^2]\cong \mathbb{C}[\mathbb{C}\oplus\mathbb{C}^*]$? –  math-visitor Jul 11 '12 at 4:23
It is because the action you defined $GL(n)$ act on $(\mathbb{C}^n)^*\cong \mathbb{C}^n$ by inverse transpose. –  Hoxide Jul 11 '12 at 4:26
Do you mean $(\mathbb{C}^n)^*$ to be the dual of $\mathbb{C}^n$, while you mean $C^*$ to mean the units in $C$? –  math-visitor Jul 11 '12 at 4:27
I'm sorry for the confusion. Yes $\mathbb{C^n}^*$ is the dual. –  Hoxide Jul 11 '12 at 4:30
Thank you! =) But another quick question: why did you write $\mathbb{C}^n \oplus (\mathbb{C}^n)^*$, rather than $\mathbb{C}^n \times (\mathbb{C}^n)^*$? Do they mean the same thing? –  math-visitor Jul 11 '12 at 4:32

In response to Question 3:

The Lie group $G= \mathbb{R}\times \mathbb{R}\times S^1$ where $S^1=\{ z\in\mathbb{C}: |z|=1 \}$ with group operation $$(x_1, y_1,u_1)\cdot (x_2,y_2, u_2)= (x_1 + x_2,y_1+y_2,e^{ix_1y_2}u_1 u_2)$$ is not a matrix group.

This was obtained from Brian Hall's Lie Groups, Lie Algebras, and representations.

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