Skip to main content
typo
Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 286

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of ChevaleyChevalley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit formulation.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here  .

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale Lectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevaley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit formulation.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here  .

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale Lectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevalley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit formulation.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here.

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale Lectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

added 82 characters in body
Source Link
Jim Humphreys
  • 52.9k
  • 4
  • 120
  • 240

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevaley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit statementformulation.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here .

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale lectures on Chevalley groupsLectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (and is partpart of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevaley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit statement.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person.

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's Yale lectures on Chevalley groups, republished in 2017 in typeset and edited form by AMS (and is part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevaley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit formulation.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here .

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale Lectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.

Source Link
Jim Humphreys
  • 52.9k
  • 4
  • 120
  • 240

There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevaley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit statement.

Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.

There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person.

See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's Yale lectures on Chevalley groups, republished in 2017 in typeset and edited form by AMS (and is part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.