Recently, I'm reading V. L. Popov's paper "Picard groups of homogeneous spaces of linear algebraic groups and one-dimensional homogeneous vector bundles" V. L. Popov, 1974, and I got confused about "Theorem 6" in that paper which says that

"Let $G$ be a connected linear algebraic group with radical $R$. Then $\mathrm{Pic}(G)$ is isomorphic to the fundamental group of the semisimple group $G/R$."

This theorem follows from "Theorem 3" and "Theorem 4" in that paper.

However, this seems to contradict the following example. Consider $G = \mathrm{GL}(n,\mathbb{C})$, whose radical $R$ is isomorphic to $\mathbb{G}_{m}$. Then the homogeneous space is $G/R = \mathrm{PGL}(n,\mathbb{C})$. We know that $\mathrm{Pic}(\mathrm{GL}(n,\mathbb{C}))$ is $0$, but $\pi_{1}(\mathrm{PGL}(n,\mathbb{C}))=\mathbb{Z}/n\mathbb{Z}$. Can anyone explain what I've understood incorrectly?

  • $\begingroup$ I'm not saying this is wrong, I'm just curious: Why doesn't $\mathrm{GL}_n$ have non-trivial line bundles? $\endgroup$
    – Ben
    Nov 5, 2016 at 0:01
  • 2
    $\begingroup$ The reason is that $\mathrm{GL}(n,\mathbb{C})$ is the complement of the determinant variety in $M_{n}(\mathbb{C})=\mathbb{A}^{n^{2}}$, the affine space of all $n$-by-$n$ matrices, so we have the following short exact sequence [Hartshorne, P.133] $$\mathbb{Z} \rightarrow \mathrm{Pic}(\mathbb{A}^{n^{2}}) \rightarrow \mathrm{Pic}(\mathrm{GL}(n,\mathbb{C})) \rightarrow 0.$$ Then this forces $\mathrm{Pic}(GL(n,\mathbb{C}))$ to be $0$, since the middle term is trivial. $\endgroup$
    – Shuai
    Nov 5, 2016 at 1:32
  • $\begingroup$ Good point, complement to a divisor in $\mathbb{A}$, thank you. $\endgroup$
    – Ben
    Nov 5, 2016 at 1:39
  • $\begingroup$ I copied the theorem word by word. I also read the proof, it seems ok for me. I think "Theorem 3" says that "For every connected linear group $G$ there exists a central isogeny $\pi: \widetilde{G}\rightarrow G$ such that $Pic(G)=0$, and the group $Ker(\pi)$ is isomorphic to the fundamental group of the semisimple group $G/R$, where $R$ is the radical of $G$." Then we can apply the theorem to where $G=GL(n,\mathbb{C})$, then we have a short exact sequence $\endgroup$
    – Shuai
    Nov 5, 2016 at 5:56
  • $\begingroup$ (Sry, in "Theorem 3" above, it should be $Pic(\widetilde{G})=0$) $$\hat{\widetilde{G}}\rightarrow \hat{ker(\pi)}\rightarrow Pic(G/R)\rightarrow 0$$ It's proved the first map factors through the commutator subgroup of $\widetilde{G}$, we get "Theorem 6"... $\endgroup$
    – Shuai
    Nov 5, 2016 at 6:07

1 Answer 1


This is really just a long comment since I can't access the paper.

There definitely seems to be something wrong with the formula, but formulas that I do know are correct basically perfectly account for the missing factor.

Namely, here are two facts that I definitely know. Let me denote by $X^\ast(G)$ the character group $\text{Hom}(G,\mathbf{G}_m)$.

Fact 1: Let $\varphi:G'\to G$ be a map of connected algebraic groups with $\ker\varphi$ of multiplicative type (i.e. $\ker\varphi$ is geometrically a product of tori and roots of unity). Then, there is the following exact sequence: $$0\to X^\ast(G)\to X^\ast(G')\to X^\ast(\ker\varphi)\to \text{Pic}(G)\to\text{Pic}(G')\to 0$$


Fact 2: If $X^\ast(G)=0$ (and $G$ is a connected algebraic group) then $G$ has a universal cover, and $\text{Pic}(G)=X^\ast(\pi_1(G))$.

So, now, suppose, for example, that $G$ is a reductive group. Then, of course, we know that $R(G)$ is a torus and so $\varphi:G\to G/R(G)$ has multiplicative kernel. So, applying Fact 1 we get

$$0\to X^\ast(G/R(G))\to X^\ast(G)\to X^\ast(R(G))\to \text{Pic}(G/R(G))\to \text{Pic}(G)\to 0$$

Now, since $G/R(G)$ is semi-simple, we have that $X^\ast(G/R(G))=0$ (indeed, the image of $G/R(G)$ in $\mathbf{G}_m$ would be connected, semisimple, and abelian--so trivial). Thus, the above really reduces to

$$0\to X^\ast(G)\to X^\ast(R(G))\to \text{Pic}(G/R(G))\to\text{Pic}(G)\to 0$$

Then, using Fact 2, and the just mentioned fact that $X^\ast(G/R(G))=0$, we have that $\text{Pic}(G/R(G))=X^\ast(\pi_1(G/R(G)))$. So, finally, our sequence looks like

$$0\to X^\ast(G)\to X^\ast(R(G))\to X^\ast(\pi_1(G))\to \text{Pic}(G)\to 0$$

Now, let's run this for $G=\text{PGL}_n$ so that $R(G)=\mathbf{G}_m$, and $G/R(G)=\text{PGL}_n$. Then, evidently:




Thus, it remains to find what $X^\ast(\pi_1(\text{PGL}_n))$ is. But, $\text{PGL}_n=\text{PSL}_n$ and $\text{SL}_n\to\text{PGL}_n$ is a central isogeny with kernel $\mu_n$. Since $\text{SL}_n$ is simply connected, this implies that $\pi_1(\text{PGL}_n)=\mu_n$. Thus,


Thus, we have, finally, the sequence

$$0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}\to \text{Pic}(\text{GL}_n)\to 0$$

which implies the desired result (that $\text{Pic}(\text{GL}_n)=0$) since the map $X^\ast(\text{GL}_n)\to X^\ast(\mathbf{G}_m)$ is multiplication by $n$ (since any $\varphi\in X^\ast(\text{GL}_n)$ factors through the determinant which, essentially, is multiplication by $n$ on $\mathbf{G}_m$: the composition $\mathbf{G}_m\xrightarrow{\approx}R(\text{GL}_n)\xrightarrow{\det}\mathbf{G}_m$ is multiplication by $n$).

In general, since $\text{Pic}(G/R(G))=X^\ast(\pi_1(G/R(G))$ and $\pi_1(G/R(G))$ is some finite abelian group, we have, non-canonically that $X^\ast(\pi_1(G/R(G))=\pi_1(G/R(G))$. Thus, the above analysis shows that we have a short exact sequence

$$0\to X^\ast(G)\to X^\ast(R(G))\to \pi_1(G/R(G)))\to \text{Pic}(G)\to 0$$

Thus, since $\pi_1(G/R(G))$ and $\text{Pic}(G)$ are finite groups, if they are isomorphic then the map $\pi_1(G/R(G))\to \text{Pic}(G)$ is an isomorphism (since they're of the same order and finite) and thus $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism. Conversely, if $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism, then evidently $\pi_1(G/R(G))\to \text{Pic}(G)$ is an ismorphism.

So, now note that the morphism $X^\ast(G/D(G))\to X^\ast(G)$ is evidently an isomorphism. Thus, we see that $X^\ast(G)\to X^\ast(R(G))$ is an isomorphism if and only if $X^\ast(G/D(G))\to X^\ast(R(G))$ is an isomorphism. But, note that since $G/D(G)$ and $R(G)$ are tori, this is equivalent to the statement that $R(G)\to G/D(G)$ is an isomorphism. But, $R(G)=Z(G)^\circ$ (we can ignore reduced subschemes because we're in characteristic $0$) and it's well known that $G=D(G)R(G)$ and thus $R(G)\to G/D(G)$ is surjective (which implies the claim above that $X^\ast(G)\to X^\ast(R(G))$ is injective) and it's an isomorphism if and only if $R(G)\cap D(G)$ is trivial.

So, the upshot of all of this is the following:

Conclusion: There is a canonical surjection $\pi_1(G/R(G))\to \text{Pic}(G)$ which is an isomorphism if and only if $R(G)\cap D(G)$ is trivial. In fact, the kernel of this map has size $|R(G)\cap D(G)|$. In fact, since $\pi_1(G/R(G))$ and $\text{Pic}(G)$ are finite abelian groups, they are isomorphic if and only if $\pi_1(G/R(G))\to \text{Pic}(G)$ is an isomorphism. Thus, $\pi_1(G/R(G))$ is isomorphic to $\text{Pic}(G)$ if and only if $D(G)\cap R(G)$ is trivial.

I haven't thought too deeply about when $R(G)\cap D(G)$ is non-trivial--do you know an example in which $G$ is not already semisimple? Anyways, it seems that the stated theorem is wrong precisely because $R(\text{GL}_n)\cap D(\text{GL}_n)$ is non-trivial, and exactly order $n$ accounting for your $\mathbb{Z}/n\mathbb{Z}$ discrepancy.

  • $\begingroup$ Maybe the example $G=PGL(n,\mathbb{C})\times \mathbb{G}_{m}$ works? SInce $\mathbb{G}_{m}$ is rational, we know $Pic(G)=Pic(PGL(n,\mathbb{C}))\times Pic(\mathbb{G}_{m})=\mathbb{Z}/n\mathbb{Z}$, $\pi_{1}(PGL(n,\mathbb{C}))=\mathbb{Z}/n\mathbb{Z}$, $G$ is not semisimple, $D(G)\cap R(G)=0$ $\endgroup$
    – Shuai
    Nov 5, 2016 at 20:21
  • $\begingroup$ Thanks a lot for your answer! And I know the decomposition $G=D(G)R(G)$ is true for reductive groups, is it true for any connected algebraic group? $\endgroup$
    – Shuai
    Nov 5, 2016 at 20:27
  • $\begingroup$ @Shuai Ah, that's a good example! That's a good question about whether $G=D(G)R(G)$ always. I certainly only usually think about it for $G$ reductive, but if I think of the proof I don't initially see why it doesn't apply for arbitrary $G$. Namely, $G/R(G)D(G)$ is simultaneously semisimple (it's a quotiento f $G/R(G)$ which is semisimple), abelian (it's a quotient of $G/D(G)$), and connected (it's a quotient of $G$), so trivial. I guess the related fact that SHOULDN'T be true, is the stronger claim that $G=D(G)Z(G)$. $\endgroup$ Nov 5, 2016 at 21:46
  • $\begingroup$ If $G$ is reductive then $Z(G)$ is finite index in $R(G)$ (it's its connected component group) and thus $G/D(G)Z(G)$ is finite over $G/D(G)R(G)$ which is trivial, so $G/D(G)Z(G)$ is finite. But, $G$ is connected thus so is $G/D(G)Z(G)$ which implies that it's trivial (again, assuming that we're working over an algebraically closed field of characteristic $0$--this should extend to any field of characteristic $0$ by replacing things with `geometrically connected'). Anyways, I do not think that this $G=Z(G)D(G)$ is true for arbitrary connected $G$. Is it even true for $B_n$ (the standard $\endgroup$ Nov 5, 2016 at 21:48
  • $\begingroup$ Borel in $\text{GL}_n$)? Anyways, I guess it's also somewhat clear that $G=D(G)R(G)$ should hold true generally since it's true for $G$ reductive, and its true for $G$ unipotent, and it probably passes over extensions in which case one can consider $1\to R_u(G)\to G\to G/R_u(G)\to 1$. Anyways, sorry for the mealymouthed answers--I'm not an expert on algebraic groups. $\endgroup$ Nov 5, 2016 at 21:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .