# Union of all Borel subgroups is the whole group

Let $$G$$ be a connected algebraic group. Then the union of all of its Borel subgroups is $$G$$ itself. I am following this proof on page 70 lemma 26.3.

Let's just focus on the first part proving the union of all Borel subgroups is closed. But I found the proof is a kinda flawed. Let $$B$$ be a Borel subgroup. Consider $$G\times B\xrightarrow{\phi} G\times G\xrightarrow{\pi} G/B\times G\xrightarrow{\gamma} G$$ where $$\phi:(g,b)\mapsto(g,gbg^{-1})$$, $$\pi:(g,h)\mapsto(gB,h)$$, $$\gamma:(gB,h)\mapsto h$$.

In total, the image is the union of the conjugates of the Borel subgroup. We would like to show it is closed. Then we want to show $$im(\pi\phi)$$ is closed then by $$G/B$$ projectivity, we could conclude. But then I ran into trouble of being convinced by the proof.

First, he wants to show $$im(\phi)$$ is closed. He cited a result proved before, corollary 16.5, which claims that the image of an algebraic group homomorphism is an algebraic group (closed). But this $$\phi$$ is not a group homomorphism, because $$\phi((g,h)(x,y))=\phi(gx,hy)=(gx,gxhyx^{-1}g^{-1})$$ but $$\phi((g,h))\phi((x,y))=(g,ghg^{-1})(x,xyx^{-1})=(gx,ghg^{-1}xyx^{-1})$$. Apparently, they are not equal...

Second, even if assume we have got $$im(\phi)$$ is closed, we want the image under $$\pi$$ is closed but $$\pi$$ is an open (surjective) map which was also proved. However, I do not think open surjective map implies closed map.

Consider the map $$G\times G\to G\times G$$ given by $$(g,g')\mapsto (g,gg'g^{-1})$$. This is an isomorphism with inverse given by $$(x,y)\mapsto (x,x^{-1}yx)$$. So, since $$G\times B$$ is a closed subset of $$G\times G$$, the image in $$G\times G$$ is closed.

Why the image of $$\mathrm{im}(\phi)$$ under $$\pi$$ is closed is slightly more sophisticated. Namely, note that $$(G/B)\times G$$ is just $$(G\times G)/(B\times 1)$$. Now, since $$G\times G\to (G/B)\times G$$ is a quotient map, to check a subset of $$(G/B)\times G$$ is closed it suffices to check the preimage is closed in $$G\times G$$. But, $$\mathrm{im}(\phi)$$ is quickly checked to be $$B\times 1$$ stable, so $$\pi^{-1}(\pi(\mathrm{im}(\phi)))=\mathrm{im}(\phi)$$.

NB: I'm using classical language since this is what Szamuely is using. Let me know if you need it translated into schemes.

• Thank you! Last question why it is enough to check the preimage is closed for $G\times G\to G/B\times G$? The map is a quotient map: open surjective continuous. But somehow I have in my mind that an open surjective continuous map need not be closed? Say $\pi:X\to Y$ is open surjective continuous. $U$ is closed, if $\pi^{-1}(U)$ is closed implies $U$ is closed for any $U$. Does it implies that it is a closed map?
– CO2
Aug 15, 2020 at 22:55
• Because a subset of a quotient space is closed iff its preimage is closed. That's not the same as being a closed map. Aug 15, 2020 at 22:58
• Ah you are right. We learnt $U$ is open if and only if $\pi^{-1}U$ is open for a quotient map. But the pullback commutes with complement So also true for closed sets. It is always a bit confusing when it relates to open and closed maps... Also I think it would be nice to see the scheme way. I am learning them right now. Thank you.
– CO2
Aug 15, 2020 at 23:04