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Let $k$ be a field. Consider the functor

$F : \mathrm{Alg}(k) \to \mathrm{Set}, ~ R \mapsto \mathrm{GL}_n(R) / R^*$

[You might call this $\mathrm{PGL}_n^{\text{naive}}$, since it does not coincide on the correct $\mathrm{PGL}_n = \mathrm{GL}_n / \mathbb{G}_m$ - only on local $k$-algebras. See also Milne's script on algebraic groups, Example I.9.5. But this is not really relevant for this question]

I claim that $F$ is representable by Freyd's representability criterion (Mac Lane, Categories for the Working Mathematician, Theorem V.6.3): It is easy to verify that $F$ preserves all limits. Every solution set for $\mathrm{GL}_n$ is also one for $F$, but $\mathrm{GL}_n$ is even representable.

Question. Which specific $k$-algebra represents $F$?

Edit. I think I can spell out the proof of the Theorem in this special case: Consider the category $\int F$ of elements of $F$: Objects are pairs $(R,s)$, where $R$ is a $k$-algebra and $s \in F(R)$. A morphism $(R,s) \to (R',s')$ is a map of $k$-algebras $R \to R'$ which maps $s \mapsto s'$. Now $\int F$ is complete (since $F$ is continuous). Besides, it has a weakly initial object, namely the representation of $\mathrm{GL}_n$, that is $w=(k[\{X_{ij}\}]_{\mathrm{det}},\overline{(X_{ij})})$. I hope the notation is clear. Now the proof of Theorem V.6.1 tells us how to construct an initial object: We have to take the equalizer $v$ of all endomorphisms of $w$ in $\int F$. This means that $v=(R,\overline{(X_{ij})})$, where $R \subseteq k[\{X_{ij}\}]_{\mathrm{det}}$ is the subalgebra whose elements are fix under every $k$-algebra homomorphism $k[\{X_{ij}\}]_{\mathrm{det}} \to k[\{X_{ij}\}]_{\mathrm{det}}$, $X_{ij} \mapsto P_{ij}$, such that $\overline{(P_{ij})} = \overline{(X_{ij})}$ in $\mathrm{PGL}_n$, i.e. $P_{ij} = \lambda X_{ij}$ for some $\lambda \in (k[\{X_{ij}\}]_{\mathrm{det}})^* = k^* \cdot \mathrm{det}^{\mathbb{Z}}$ (here we use that $\mathrm{det}$ is an irreducible polynomial).

In other words, $R$ consists of those polynomials in the $X_{ij}$ localized at $\mathrm{det}$ such that the substitution $X_{ij} \mapsto \lambda \cdot X_{ij}$ for some $\lambda \in k^*$, and also $X_{ij} \mapsto \mathrm{det} \cdot X_{ij}$, doesn't change them. Is there any way to make this algebra more explicit? Can be describe it by generators and relations?

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Since $\textrm{PGL}_n(R)$ is the quotient of $\textrm{GL}_n(R)$ by the action of $R^*$, one expects the corresponding $k$-algebra to be the invariants under $R^*$? But that can't possibly be right. Could you perhaps explain in more detail why the conditions of the special adjoint functor theorem are satisfied? It is not obvious to me that $\textrm{Alg}(k)$ has a coseparating set. –  Zhen Lin Mar 16 '12 at 21:02
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Thanks for the comment about the coseparating set; now I've proven it with Freyd's representability criterion. // Meanwhile I've realized that the proof can be made mor explicit and indeed we get the $k^*$-invariants of $k[\{X_{ij}\}]$ as an representing object. –  Martin Brandenburg Mar 18 '12 at 16:16
    
Glad I could help. I guess, when $k$ is infinite, they must be the homogeneous polynomials of degree a multiple of $n$, divided by an appropriate power of the determinant. –  Zhen Lin Mar 18 '12 at 17:31

2 Answers 2

Argh! The premise of my question

It is easy to verify that $F$ preserves all limits.

is wrong. It is correct that $F$ preserves products, but equalizers are not preserved; see below. And this is the reason why the whole search for a representation issue doesn't make sense.

Let $f,g : R \to S$ be two homomorphisms of algebras. Their equalizer is $E = \{r \in R : f(r)=g(r)\}$. The equalizer of $F(f),F(g) : F(R) \to F(S)$ consists of those $[M]$, $M \in \mathrm{GL}_n(R)$, such that $g(M)^{-1} f(M) \subseteq S^* \subseteq \mathrm{GL}_n(S)$. But the image of $F(E) \to F(R)$ consists of those $[M]$, $M \in \mathrm{GL}_n(R)$, such that there is some $r \in R^*$ such that $r M \in \mathrm{GL}_n(E)$, which is easily seen to be equivalent to $g(M)^{-1} f(M) = g(r)^{-1} f(r)$ for some $r \in R^*$. Thus, if $R=S$, $g=\mathrm{id}$, on the one hand we have $f(M) \subseteq S^*$ and on the other hand $f(M) \in f(R^*)$. These conditions are different for $n \geq 2$. Perhaps someone can add an example.

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We have that $PGL(n) \cong PSL(n) \rtimes G_m / G_m^n$! So the algebra representing $PGL(n)$ is a crossed product of the one representing $PSL(n)$ by the one representing $G_m / G_m^n$. I am not sure, if the representing algebra of $PSL(n)$ is easier to construct than that for $PGL(n)$, but the center of $SL(n)$ is finite.

The semidirect product comes from the exact sequence $$ SL(n) \rightarrow det: GL(n) \rightarrow G_m \rightarrow G_m / G_m^n,$$ and consequently $$ PSL(n) \rightarrow PGL(n) \rightarrow G_m / G_m^n.$$

Here, $G_m$ is the mutiplicative group and $G_m^n$ the subgroup of $n$-th powers, the representing polynomial equation of the quotient is $x^n=1$. Sometimes the quotient is called the group of $n$-th roots of unity.

Writing down the representing algebra of the group schemes is beyond my experience and will take me to much time. So perhaps this is just a comment rather than an answer.

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How do you prove your isomorphism? –  darij grinberg Mar 29 '12 at 12:27
    
Dito, and can you make this crossed product more explicit? What is the representing algebra of $G_m / G_m^n$? –  Martin Brandenburg Mar 29 '12 at 13:04

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