CW-Structure of Spin(n)
$begingroup$
I cannot find any information about the CW-Structure of $Spin(n)$ groups. Clearly $pi_1=pi_2=0$ and I think $H_3(Spin(n))=mathbb Z$ $(ngeq 5)$, so the $3$-skeleton is $S^3$. What is the $4$-skeleton?
algebraic-topology homotopy-theory
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
$endgroup$
add a comment |
$begingroup$
I cannot find any information about the CW-Structure of $Spin(n)$ groups. Clearly $pi_1=pi_2=0$ and I think $H_3(Spin(n))=mathbb Z$ $(ngeq 5)$, so the $3$-skeleton is $S^3$. What is the $4$-skeleton?
algebraic-topology homotopy-theory
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
$endgroup$
add a comment |
$begingroup$
I cannot find any information about the CW-Structure of $Spin(n)$ groups. Clearly $pi_1=pi_2=0$ and I think $H_3(Spin(n))=mathbb Z$ $(ngeq 5)$, so the $3$-skeleton is $S^3$. What is the $4$-skeleton?
algebraic-topology homotopy-theory
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
$endgroup$
I cannot find any information about the CW-Structure of $Spin(n)$ groups. Clearly $pi_1=pi_2=0$ and I think $H_3(Spin(n))=mathbb Z$ $(ngeq 5)$, so the $3$-skeleton is $S^3$. What is the $4$-skeleton?
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 12:43
Wilhelm L.Wilhelm L.
41117
41117
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each $n$ there is a fibration sequence
$$Spin_nrightarrow Spin_{n+1}rightarrow S^n$$ covering the corresponding fibration sequence $SO_nrightarrow SO_{n+1}rightarrow S^n$. The point is that the inclusion $Spin_nhookrightarrow Spin_{n+1}$ is $(n-1)$-connected, so the $4$-skeleton of $Spin_n$ is the same as the $4$-skeleton of $Spin_6$ for each $ngeq 6$.
Now $Spin_5cong Sp_2$ and it is know that $Sp_2simeq S^3cup_{nu'} e^7cup e^{10}$, so the $4$-skeleton of $Spin_5$. Of course $Spin_3cong S^3$ and $Spin_4cong S^3times S^3$ which have $4$-skeletons $S^3$, $S^3vee S^3$ and $S^3$, respectively.
Also $Spin_6cong SU_4$ and the $5$-skeleton of $SU_4$ is $Sigma mathbb{C}P^2simeq S^3cup_{eta_3}e^5$, so by the previous comments, in fact the $5$-skeleton of $Spin_n$ for $ngeq 6$ is $S^3cup_{eta_3}e^5$.
We also have a fibration $G_2rightarrow Spin_7rightarrow S^7$, and it's not that difficult to see that $Spin_7simeq S^3cup_{eta_3}cup e^6cup e^6cupdots$, which gives the $6$-sekelton of $Spin_n$ for $ngeq 7$. In fact, the previous fibration splits when localised away from $2$ so there is a $frac{1}{2}$-local homotopy equivalance $Spin_7simeq G_2times S^7$. Finally we mention that there is a homeomorphism $Spin_8cong Spin_7times S^7$, so you get the $7$-skeleton of $Spin_8$ from that of $Spin_7$ wedged with an extra $7$-cell. I'm not aware of any further isomorphisms for the higher rank spinor groups.
A full CW decomposition was given by Araki in his paper On the homology of spinor groups, which was predated by some related results of Borel in Sur I'homologie et la cohomologie des groupes de Lie compacts connexes. As you might expect, the results tend to get a bit complicated as the rank grows, and whilst Araki calculates explicit attaching maps, he does not explicitly calculate their homotopy classes, so depending on what information you are looking for, you may be more interested in his homological calculations. And to this end I should refer you to the paper The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$ of Pittie, which does exactly what it says.
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
$endgroup$
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
|
show 7 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084420%2fcw-structure-of-spinn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For each $n$ there is a fibration sequence
$$Spin_nrightarrow Spin_{n+1}rightarrow S^n$$ covering the corresponding fibration sequence $SO_nrightarrow SO_{n+1}rightarrow S^n$. The point is that the inclusion $Spin_nhookrightarrow Spin_{n+1}$ is $(n-1)$-connected, so the $4$-skeleton of $Spin_n$ is the same as the $4$-skeleton of $Spin_6$ for each $ngeq 6$.
Now $Spin_5cong Sp_2$ and it is know that $Sp_2simeq S^3cup_{nu'} e^7cup e^{10}$, so the $4$-skeleton of $Spin_5$. Of course $Spin_3cong S^3$ and $Spin_4cong S^3times S^3$ which have $4$-skeletons $S^3$, $S^3vee S^3$ and $S^3$, respectively.
Also $Spin_6cong SU_4$ and the $5$-skeleton of $SU_4$ is $Sigma mathbb{C}P^2simeq S^3cup_{eta_3}e^5$, so by the previous comments, in fact the $5$-skeleton of $Spin_n$ for $ngeq 6$ is $S^3cup_{eta_3}e^5$.
We also have a fibration $G_2rightarrow Spin_7rightarrow S^7$, and it's not that difficult to see that $Spin_7simeq S^3cup_{eta_3}cup e^6cup e^6cupdots$, which gives the $6$-sekelton of $Spin_n$ for $ngeq 7$. In fact, the previous fibration splits when localised away from $2$ so there is a $frac{1}{2}$-local homotopy equivalance $Spin_7simeq G_2times S^7$. Finally we mention that there is a homeomorphism $Spin_8cong Spin_7times S^7$, so you get the $7$-skeleton of $Spin_8$ from that of $Spin_7$ wedged with an extra $7$-cell. I'm not aware of any further isomorphisms for the higher rank spinor groups.
A full CW decomposition was given by Araki in his paper On the homology of spinor groups, which was predated by some related results of Borel in Sur I'homologie et la cohomologie des groupes de Lie compacts connexes. As you might expect, the results tend to get a bit complicated as the rank grows, and whilst Araki calculates explicit attaching maps, he does not explicitly calculate their homotopy classes, so depending on what information you are looking for, you may be more interested in his homological calculations. And to this end I should refer you to the paper The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$ of Pittie, which does exactly what it says.
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
$endgroup$
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
|
show 7 more comments
$begingroup$
For each $n$ there is a fibration sequence
$$Spin_nrightarrow Spin_{n+1}rightarrow S^n$$ covering the corresponding fibration sequence $SO_nrightarrow SO_{n+1}rightarrow S^n$. The point is that the inclusion $Spin_nhookrightarrow Spin_{n+1}$ is $(n-1)$-connected, so the $4$-skeleton of $Spin_n$ is the same as the $4$-skeleton of $Spin_6$ for each $ngeq 6$.
Now $Spin_5cong Sp_2$ and it is know that $Sp_2simeq S^3cup_{nu'} e^7cup e^{10}$, so the $4$-skeleton of $Spin_5$. Of course $Spin_3cong S^3$ and $Spin_4cong S^3times S^3$ which have $4$-skeletons $S^3$, $S^3vee S^3$ and $S^3$, respectively.
Also $Spin_6cong SU_4$ and the $5$-skeleton of $SU_4$ is $Sigma mathbb{C}P^2simeq S^3cup_{eta_3}e^5$, so by the previous comments, in fact the $5$-skeleton of $Spin_n$ for $ngeq 6$ is $S^3cup_{eta_3}e^5$.
We also have a fibration $G_2rightarrow Spin_7rightarrow S^7$, and it's not that difficult to see that $Spin_7simeq S^3cup_{eta_3}cup e^6cup e^6cupdots$, which gives the $6$-sekelton of $Spin_n$ for $ngeq 7$. In fact, the previous fibration splits when localised away from $2$ so there is a $frac{1}{2}$-local homotopy equivalance $Spin_7simeq G_2times S^7$. Finally we mention that there is a homeomorphism $Spin_8cong Spin_7times S^7$, so you get the $7$-skeleton of $Spin_8$ from that of $Spin_7$ wedged with an extra $7$-cell. I'm not aware of any further isomorphisms for the higher rank spinor groups.
A full CW decomposition was given by Araki in his paper On the homology of spinor groups, which was predated by some related results of Borel in Sur I'homologie et la cohomologie des groupes de Lie compacts connexes. As you might expect, the results tend to get a bit complicated as the rank grows, and whilst Araki calculates explicit attaching maps, he does not explicitly calculate their homotopy classes, so depending on what information you are looking for, you may be more interested in his homological calculations. And to this end I should refer you to the paper The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$ of Pittie, which does exactly what it says.
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
$endgroup$
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
|
show 7 more comments
$begingroup$
For each $n$ there is a fibration sequence
$$Spin_nrightarrow Spin_{n+1}rightarrow S^n$$ covering the corresponding fibration sequence $SO_nrightarrow SO_{n+1}rightarrow S^n$. The point is that the inclusion $Spin_nhookrightarrow Spin_{n+1}$ is $(n-1)$-connected, so the $4$-skeleton of $Spin_n$ is the same as the $4$-skeleton of $Spin_6$ for each $ngeq 6$.
Now $Spin_5cong Sp_2$ and it is know that $Sp_2simeq S^3cup_{nu'} e^7cup e^{10}$, so the $4$-skeleton of $Spin_5$. Of course $Spin_3cong S^3$ and $Spin_4cong S^3times S^3$ which have $4$-skeletons $S^3$, $S^3vee S^3$ and $S^3$, respectively.
Also $Spin_6cong SU_4$ and the $5$-skeleton of $SU_4$ is $Sigma mathbb{C}P^2simeq S^3cup_{eta_3}e^5$, so by the previous comments, in fact the $5$-skeleton of $Spin_n$ for $ngeq 6$ is $S^3cup_{eta_3}e^5$.
We also have a fibration $G_2rightarrow Spin_7rightarrow S^7$, and it's not that difficult to see that $Spin_7simeq S^3cup_{eta_3}cup e^6cup e^6cupdots$, which gives the $6$-sekelton of $Spin_n$ for $ngeq 7$. In fact, the previous fibration splits when localised away from $2$ so there is a $frac{1}{2}$-local homotopy equivalance $Spin_7simeq G_2times S^7$. Finally we mention that there is a homeomorphism $Spin_8cong Spin_7times S^7$, so you get the $7$-skeleton of $Spin_8$ from that of $Spin_7$ wedged with an extra $7$-cell. I'm not aware of any further isomorphisms for the higher rank spinor groups.
A full CW decomposition was given by Araki in his paper On the homology of spinor groups, which was predated by some related results of Borel in Sur I'homologie et la cohomologie des groupes de Lie compacts connexes. As you might expect, the results tend to get a bit complicated as the rank grows, and whilst Araki calculates explicit attaching maps, he does not explicitly calculate their homotopy classes, so depending on what information you are looking for, you may be more interested in his homological calculations. And to this end I should refer you to the paper The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$ of Pittie, which does exactly what it says.
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
$endgroup$
For each $n$ there is a fibration sequence
$$Spin_nrightarrow Spin_{n+1}rightarrow S^n$$ covering the corresponding fibration sequence $SO_nrightarrow SO_{n+1}rightarrow S^n$. The point is that the inclusion $Spin_nhookrightarrow Spin_{n+1}$ is $(n-1)$-connected, so the $4$-skeleton of $Spin_n$ is the same as the $4$-skeleton of $Spin_6$ for each $ngeq 6$.
Now $Spin_5cong Sp_2$ and it is know that $Sp_2simeq S^3cup_{nu'} e^7cup e^{10}$, so the $4$-skeleton of $Spin_5$. Of course $Spin_3cong S^3$ and $Spin_4cong S^3times S^3$ which have $4$-skeletons $S^3$, $S^3vee S^3$ and $S^3$, respectively.
Also $Spin_6cong SU_4$ and the $5$-skeleton of $SU_4$ is $Sigma mathbb{C}P^2simeq S^3cup_{eta_3}e^5$, so by the previous comments, in fact the $5$-skeleton of $Spin_n$ for $ngeq 6$ is $S^3cup_{eta_3}e^5$.
We also have a fibration $G_2rightarrow Spin_7rightarrow S^7$, and it's not that difficult to see that $Spin_7simeq S^3cup_{eta_3}cup e^6cup e^6cupdots$, which gives the $6$-sekelton of $Spin_n$ for $ngeq 7$. In fact, the previous fibration splits when localised away from $2$ so there is a $frac{1}{2}$-local homotopy equivalance $Spin_7simeq G_2times S^7$. Finally we mention that there is a homeomorphism $Spin_8cong Spin_7times S^7$, so you get the $7$-skeleton of $Spin_8$ from that of $Spin_7$ wedged with an extra $7$-cell. I'm not aware of any further isomorphisms for the higher rank spinor groups.
A full CW decomposition was given by Araki in his paper On the homology of spinor groups, which was predated by some related results of Borel in Sur I'homologie et la cohomologie des groupes de Lie compacts connexes. As you might expect, the results tend to get a bit complicated as the rank grows, and whilst Araki calculates explicit attaching maps, he does not explicitly calculate their homotopy classes, so depending on what information you are looking for, you may be more interested in his homological calculations. And to this end I should refer you to the paper The integral homology and cohomology rings of $SO(n)$ and $Spin(n)$ of Pittie, which does exactly what it says.
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
edited Jan 23 at 15:17
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
answered Jan 23 at 14:01
TyroneTyrone
4,92511225
4,92511225
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
|
show 7 more comments
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
Real dumb question: why is the $4$-skeleton given using cells up to 5 dimensions? Also, how does one see that the $4$-skelton is $Sigma mathbb{C}P^2$? (I have heard that $Sq^2:H^3(SU(n);mathbb{Z}_2)rightarrow H^5(SU(n);mathbb{Z}_2)$ is an isomorphism (for large enough $n$) but never new a proof. Computing the $4$-skeleton would give me a proof!)
$endgroup$
– Jason DeVito
Jan 23 at 14:48
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
I think that that was a typo and the $4$-skeleton of $Spin(n)$ for $ngeq 6$ is just $S^3$.
$endgroup$
– Wilhelm L.
Jan 23 at 14:54
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, yeah, that was just a typo, that's really the 5-skeleton, which is where the next cell lives, so it kinda seemed relevant to include that information. Thanks for the correction, Wilhelm.
$endgroup$
– Tyrone
Jan 23 at 15:08
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@JasonDeVito, There is a map $Sigmamathbb{C}P^3rightarrow SU_4$ given by sending a complex line to the generalised reflection through it (modulo a bit of fiddling to sort out the $det=1$ condition). This map is exactly how the CW structure of $SU_4$ is defined by Yokota and Whitehead, and I can try to dig out some similar references to the above if you are interested. Basically, if you follow Hatcher's discussion of the cohomology of $SO_n$, the same ideas go through to the complex case.
$endgroup$
– Tyrone
Jan 23 at 15:22
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
$begingroup$
@Tyrone: (and Wilhelm). Thanks. I do very little pure homotopy theory, so figured I was just messing up something little. And no reason to dig up any references - I am ok with blackboxing certain results. I will mess around a bit with your suggested map $Sigma mathbb{C}P^3rightarrow SU(4)$ and see what I can prove. (And, should the same argument work for $Sigma mathbb{C}P^2rightarrow SU(3)$?)
$endgroup$
– Jason DeVito
Jan 23 at 15:25
|
show 7 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084420%2fcw-structure-of-spinn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
