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Consequences from Bott Periodicity

2

1

$begingroup$

I have some questions about some arguments used in the discussion about consequnces of Bott periodity in in A Concise Course in Algebraic Topology by P. May at page 207. Her the excerpt:

Following context: Denote $U:= colim_n U(n)$ the infinite unitary group induced via canonical inclusions $U(n) subset U(n+1)$.

FIRST QUESTION: Why is the loop space $Omega BU := Hom(S^1, BU)$ $H$- equivalent to $U$?

We know that $BU = E/U$ where $E$ is the unique simply connected universal principal $U$-space and after applying $pi_0(-)$ functor we obtain $pi_0(Omega BU) = pi_1(E/U)= U$ by covering theory since $E$ is universal cover.

But does this already imply that $Omega BU $ homotopy equivalent to $U$? $pi_0$ only counts path components.

SECOND QUESTION: We know that $SU$ - the infinite special unitary group - is the universal cover of $U$ (via colimit argument). Futhermore $pi_1(U)=mathbb{Z}$.

Why does it imply that $$Omega U cong (Omega SU) times mathbb{Z}$$ as $H$-spaces (remark: $X$ is a $H$-space if there exist a "multiplication" $X times X to X$.

general-topology algebraic-topology fiber-bundles

share|cite|improve this question

edited Mar 22 at 21:50

KarlPeter

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

$endgroup$

add a comment | 

2

1

$begingroup$

I have some questions about some arguments used in the discussion about consequnces of Bott periodity in in A Concise Course in Algebraic Topology by P. May at page 207. Her the excerpt:

Following context: Denote $U:= colim_n U(n)$ the infinite unitary group induced via canonical inclusions $U(n) subset U(n+1)$.

FIRST QUESTION: Why is the loop space $Omega BU := Hom(S^1, BU)$ $H$- equivalent to $U$?

We know that $BU = E/U$ where $E$ is the unique simply connected universal principal $U$-space and after applying $pi_0(-)$ functor we obtain $pi_0(Omega BU) = pi_1(E/U)= U$ by covering theory since $E$ is universal cover.

But does this already imply that $Omega BU $ homotopy equivalent to $U$? $pi_0$ only counts path components.

SECOND QUESTION: We know that $SU$ - the infinite special unitary group - is the universal cover of $U$ (via colimit argument). Futhermore $pi_1(U)=mathbb{Z}$.

Why does it imply that $$Omega U cong (Omega SU) times mathbb{Z}$$ as $H$-spaces (remark: $X$ is a $H$-space if there exist a "multiplication" $X times X to X$.

general-topology algebraic-topology fiber-bundles

share|cite|improve this question

edited Mar 22 at 21:50

KarlPeter

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

$endgroup$

add a comment | 

$begingroup$

I have some questions about some arguments used in the discussion about consequnces of Bott periodity in in A Concise Course in Algebraic Topology by P. May at page 207. Her the excerpt:

Following context: Denote $U:= colim_n U(n)$ the infinite unitary group induced via canonical inclusions $U(n) subset U(n+1)$.

FIRST QUESTION: Why is the loop space $Omega BU := Hom(S^1, BU)$ $H$- equivalent to $U$?

We know that $BU = E/U$ where $E$ is the unique simply connected universal principal $U$-space and after applying $pi_0(-)$ functor we obtain $pi_0(Omega BU) = pi_1(E/U)= U$ by covering theory since $E$ is universal cover.

But does this already imply that $Omega BU $ homotopy equivalent to $U$? $pi_0$ only counts path components.

SECOND QUESTION: We know that $SU$ - the infinite special unitary group - is the universal cover of $U$ (via colimit argument). Futhermore $pi_1(U)=mathbb{Z}$.

Why does it imply that $$Omega U cong (Omega SU) times mathbb{Z}$$ as $H$-spaces (remark: $X$ is a $H$-space if there exist a "multiplication" $X times X to X$.

general-topology algebraic-topology fiber-bundles

share|cite|improve this question

edited Mar 22 at 21:50

KarlPeter

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

$endgroup$

share|cite|improve this question

edited Mar 22 at 21:50

KarlPeter

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

share|cite|improve this question

edited Mar 22 at 21:50

KarlPeter

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

asked Mar 22 at 20:18

KarlPeterKarlPeter

6841416

1 Answer
1

active

oldest

votes

1

$begingroup$

Consider the map of fibrations from the universal principal $U$-bundle over $BU$ to the path-loop fibration on $BU$:

$require{AMScd}$
begin{CD}
U @>>> EU @>>> BU \
@VVV @VVV @| \
Omega BU @>>> PBU @>>> BU
end{CD}

One can write down a map $EU to PBU$; thus there is a map $U to Omega BU$. Looking at the long exact sequence in homotopy and using the five-lemma, one deduces that the induced map $U to Omega BU$ is a weak homotopy equivalence because $EU simeq PBU simeq *$. (Since both source and target can be equipped with a CW structure, this can be promoted to a homotopy equivalence.)

In general, this shows $Omega BG simeq G$ for a topological group $G$. See this answer for more details.

For your second question, there is a short exact sequence of topological groups $$1 to SU to U xrightarrow{det} S^1 to 1$$ which is split, so $U cong SU rtimes S^1$ as topological groups. If we ignore the group structure and only care about the topology, then we have a homeomorphism $U cong SU times S^1$. Taking loops on both sides, we get $$Omega U cong Omega SU times Omega S^1 simeq Omega SU times mathbb{Z}$$ as loop spaces, hence as $H$-spaces.

share|cite|improve this answer

answered Mar 22 at 22:02

JHFJHF

4,9961026

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answer. One point iritates me: In the last line you claim (neglecting the left factor $Omega SU$) that $Omega S^1 simeq mathbb{Z}$. Why does it hold? I think that it only holds after passing to homtopy groups. More precisely we know that the loop space is defined as $Omega S^1 := Hom(S^1, S^1)$. On the other hand we know that for homotopy classes holds indeed $[S^1,S^1]=mathbb{Z}$ where the conection is the path component functor $pi_0(Omega S^1 )= [S^1,S^1]$. But I'm not sure why should hold $Omega S^1 simeq mathbb{Z}$ before passing to homotopy classes?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another remark: Do I understand the argument with the $H$-story correctly that already $U cong SU times S^1$ suffice since $Omega(-)$ endows in a functorial way the spaces with $H$-space structure? And therefore $H$-space property is here equivalent to functoriality of $Omega(-)$?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Omega S^1$ is homotopy discrete (it has no higher homotopy groups, because $S^1$ has no homotopy groups above degree $1$). For your second question: if $X cong Y$, then $Omega X cong Omega Y$ as loop spaces. You might be concerned about the original multiplication on $X$ and $Y$ if they were groups to begin with, but an Eckmann-Hilton argument would show that the operations coincide. Finally, note that a loop space equivalence is stronger than a $H$-space equivalence.
  $endgroup$
  – JHF
  Mar 25 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hmmm so for $Omega S^1 simeq mathbb{Z}$ you also implicitely use Whitehead, don't you? en.wikipedia.org/wiki/Whitehead_theorem Since your homotopy discreteness argument provides that $Omega S^1$ and $coprod_{z in mathbb{Z}} {*} cong mathbb{Z}$ have the same homotopy groups. Or did I misunderstood your point?
  $endgroup$
  – KarlPeter
  Mar 25 at 16:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure that works. Or you could argue $Omega S^1 simeq Omega K(mathbb{Z}, 1) simeq K(mathbb{Z}, 0) simeq mathbb{Z}$, which is basically the same argument.
  $endgroup$
  – JHF
  Mar 25 at 17:12
  

  
  
  
  

add a comment | 

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1

$begingroup$

Consider the map of fibrations from the universal principal $U$-bundle over $BU$ to the path-loop fibration on $BU$:

$require{AMScd}$
begin{CD}
U @>>> EU @>>> BU \
@VVV @VVV @| \
Omega BU @>>> PBU @>>> BU
end{CD}

One can write down a map $EU to PBU$; thus there is a map $U to Omega BU$. Looking at the long exact sequence in homotopy and using the five-lemma, one deduces that the induced map $U to Omega BU$ is a weak homotopy equivalence because $EU simeq PBU simeq *$. (Since both source and target can be equipped with a CW structure, this can be promoted to a homotopy equivalence.)

In general, this shows $Omega BG simeq G$ for a topological group $G$. See this answer for more details.

For your second question, there is a short exact sequence of topological groups $$1 to SU to U xrightarrow{det} S^1 to 1$$ which is split, so $U cong SU rtimes S^1$ as topological groups. If we ignore the group structure and only care about the topology, then we have a homeomorphism $U cong SU times S^1$. Taking loops on both sides, we get $$Omega U cong Omega SU times Omega S^1 simeq Omega SU times mathbb{Z}$$ as loop spaces, hence as $H$-spaces.

share|cite|improve this answer

answered Mar 22 at 22:02

JHFJHF

4,9961026

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answer. One point iritates me: In the last line you claim (neglecting the left factor $Omega SU$) that $Omega S^1 simeq mathbb{Z}$. Why does it hold? I think that it only holds after passing to homtopy groups. More precisely we know that the loop space is defined as $Omega S^1 := Hom(S^1, S^1)$. On the other hand we know that for homotopy classes holds indeed $[S^1,S^1]=mathbb{Z}$ where the conection is the path component functor $pi_0(Omega S^1 )= [S^1,S^1]$. But I'm not sure why should hold $Omega S^1 simeq mathbb{Z}$ before passing to homotopy classes?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another remark: Do I understand the argument with the $H$-story correctly that already $U cong SU times S^1$ suffice since $Omega(-)$ endows in a functorial way the spaces with $H$-space structure? And therefore $H$-space property is here equivalent to functoriality of $Omega(-)$?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Omega S^1$ is homotopy discrete (it has no higher homotopy groups, because $S^1$ has no homotopy groups above degree $1$). For your second question: if $X cong Y$, then $Omega X cong Omega Y$ as loop spaces. You might be concerned about the original multiplication on $X$ and $Y$ if they were groups to begin with, but an Eckmann-Hilton argument would show that the operations coincide. Finally, note that a loop space equivalence is stronger than a $H$-space equivalence.
  $endgroup$
  – JHF
  Mar 25 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hmmm so for $Omega S^1 simeq mathbb{Z}$ you also implicitely use Whitehead, don't you? en.wikipedia.org/wiki/Whitehead_theorem Since your homotopy discreteness argument provides that $Omega S^1$ and $coprod_{z in mathbb{Z}} {*} cong mathbb{Z}$ have the same homotopy groups. Or did I misunderstood your point?
  $endgroup$
  – KarlPeter
  Mar 25 at 16:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure that works. Or you could argue $Omega S^1 simeq Omega K(mathbb{Z}, 1) simeq K(mathbb{Z}, 0) simeq mathbb{Z}$, which is basically the same argument.
  $endgroup$
  – JHF
  Mar 25 at 17:12
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider the map of fibrations from the universal principal $U$-bundle over $BU$ to the path-loop fibration on $BU$:

$require{AMScd}$
begin{CD}
U @>>> EU @>>> BU \
@VVV @VVV @| \
Omega BU @>>> PBU @>>> BU
end{CD}

One can write down a map $EU to PBU$; thus there is a map $U to Omega BU$. Looking at the long exact sequence in homotopy and using the five-lemma, one deduces that the induced map $U to Omega BU$ is a weak homotopy equivalence because $EU simeq PBU simeq *$. (Since both source and target can be equipped with a CW structure, this can be promoted to a homotopy equivalence.)

In general, this shows $Omega BG simeq G$ for a topological group $G$. See this answer for more details.

For your second question, there is a short exact sequence of topological groups $$1 to SU to U xrightarrow{det} S^1 to 1$$ which is split, so $U cong SU rtimes S^1$ as topological groups. If we ignore the group structure and only care about the topology, then we have a homeomorphism $U cong SU times S^1$. Taking loops on both sides, we get $$Omega U cong Omega SU times Omega S^1 simeq Omega SU times mathbb{Z}$$ as loop spaces, hence as $H$-spaces.

share|cite|improve this answer

answered Mar 22 at 22:02

JHFJHF

4,9961026

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for answer. One point iritates me: In the last line you claim (neglecting the left factor $Omega SU$) that $Omega S^1 simeq mathbb{Z}$. Why does it hold? I think that it only holds after passing to homtopy groups. More precisely we know that the loop space is defined as $Omega S^1 := Hom(S^1, S^1)$. On the other hand we know that for homotopy classes holds indeed $[S^1,S^1]=mathbb{Z}$ where the conection is the path component functor $pi_0(Omega S^1 )= [S^1,S^1]$. But I'm not sure why should hold $Omega S^1 simeq mathbb{Z}$ before passing to homotopy classes?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another remark: Do I understand the argument with the $H$-story correctly that already $U cong SU times S^1$ suffice since $Omega(-)$ endows in a functorial way the spaces with $H$-space structure? And therefore $H$-space property is here equivalent to functoriality of $Omega(-)$?
  $endgroup$
  – KarlPeter
  Mar 23 at 0:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Omega S^1$ is homotopy discrete (it has no higher homotopy groups, because $S^1$ has no homotopy groups above degree $1$). For your second question: if $X cong Y$, then $Omega X cong Omega Y$ as loop spaces. You might be concerned about the original multiplication on $X$ and $Y$ if they were groups to begin with, but an Eckmann-Hilton argument would show that the operations coincide. Finally, note that a loop space equivalence is stronger than a $H$-space equivalence.
  $endgroup$
  – JHF
  Mar 25 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hmmm so for $Omega S^1 simeq mathbb{Z}$ you also implicitely use Whitehead, don't you? en.wikipedia.org/wiki/Whitehead_theorem Since your homotopy discreteness argument provides that $Omega S^1$ and $coprod_{z in mathbb{Z}} {*} cong mathbb{Z}$ have the same homotopy groups. Or did I misunderstood your point?
  $endgroup$
  – KarlPeter
  Mar 25 at 16:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure that works. Or you could argue $Omega S^1 simeq Omega K(mathbb{Z}, 1) simeq K(mathbb{Z}, 0) simeq mathbb{Z}$, which is basically the same argument.
  $endgroup$
  – JHF
  Mar 25 at 17:12
  

  
  
  
  

add a comment | 

$begingroup$

Consider the map of fibrations from the universal principal $U$-bundle over $BU$ to the path-loop fibration on $BU$:

$require{AMScd}$
begin{CD}
U @>>> EU @>>> BU \
@VVV @VVV @| \
Omega BU @>>> PBU @>>> BU
end{CD}

One can write down a map $EU to PBU$; thus there is a map $U to Omega BU$. Looking at the long exact sequence in homotopy and using the five-lemma, one deduces that the induced map $U to Omega BU$ is a weak homotopy equivalence because $EU simeq PBU simeq *$. (Since both source and target can be equipped with a CW structure, this can be promoted to a homotopy equivalence.)

In general, this shows $Omega BG simeq G$ for a topological group $G$. See this answer for more details.

For your second question, there is a short exact sequence of topological groups $$1 to SU to U xrightarrow{det} S^1 to 1$$ which is split, so $U cong SU rtimes S^1$ as topological groups. If we ignore the group structure and only care about the topology, then we have a homeomorphism $U cong SU times S^1$. Taking loops on both sides, we get $$Omega U cong Omega SU times Omega S^1 simeq Omega SU times mathbb{Z}$$ as loop spaces, hence as $H$-spaces.

share|cite|improve this answer

answered Mar 22 at 22:02

JHFJHF

4,9961026

$endgroup$

share|cite|improve this answer

answered Mar 22 at 22:02

JHFJHF

4,9961026

answered Mar 22 at 22:02

JHFJHF

4,9961026

answered Mar 22 at 22:02

JHFJHF

4,9961026