Mixing
Extending a monomorphism of bundles, Lemma 7.3, Atiyah, Shapiro
$begingroup$
This is from Lemma 7.2, pg17
Let $E,F$ be vector bundles on $X$ and $f:E rightarrow F$ a monomorphism on $Y$. Then if $dim F > dim E+ dim X$, $f$ can be extended to a monomorphismon on $X$ and any two such extensions are homotopic rel $Y$.
I completely do not follow the supplied proof.
It would be great if someone could provide references (or proof ) for each of the claims.
Consider the fibre bundle $Mon(E,F)$ where the fibers are the monomorphisms of $E_x rightarrow F_x$.
- This is homeomoprhic to $GL(n)/GL(n-m)$
This space is $n-m-1$ connected.
cross sections can be extended and are all homeomorphic if
$$ dim X le dim F- dim E -1 $$
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 31 at 16:55
CL.CL.
2,3102925
$endgroup$
add a comment |
$begingroup$
This is from Lemma 7.2, pg17
Let $E,F$ be vector bundles on $X$ and $f:E rightarrow F$ a monomorphism on $Y$. Then if $dim F > dim E+ dim X$, $f$ can be extended to a monomorphismon on $X$ and any two such extensions are homotopic rel $Y$.
I completely do not follow the supplied proof.
It would be great if someone could provide references (or proof ) for each of the claims.
Consider the fibre bundle $Mon(E,F)$ where the fibers are the monomorphisms of $E_x rightarrow F_x$.
- This is homeomoprhic to $GL(n)/GL(n-m)$
This space is $n-m-1$ connected.
cross sections can be extended and are all homeomorphic if
$$ dim X le dim F- dim E -1 $$
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 31 at 16:55
CL.CL.
2,3102925
$endgroup$
add a comment |
$begingroup$
This is from Lemma 7.2, pg17
Let $E,F$ be vector bundles on $X$ and $f:E rightarrow F$ a monomorphism on $Y$. Then if $dim F > dim E+ dim X$, $f$ can be extended to a monomorphismon on $X$ and any two such extensions are homotopic rel $Y$.
I completely do not follow the supplied proof.
It would be great if someone could provide references (or proof ) for each of the claims.
Consider the fibre bundle $Mon(E,F)$ where the fibers are the monomorphisms of $E_x rightarrow F_x$.
- This is homeomoprhic to $GL(n)/GL(n-m)$
This space is $n-m-1$ connected.
cross sections can be extended and are all homeomorphic if
$$ dim X le dim F- dim E -1 $$
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 31 at 16:55
CL.CL.
2,3102925
$endgroup$
This is from Lemma 7.2, pg17
Let $E,F$ be vector bundles on $X$ and $f:E rightarrow F$ a monomorphism on $Y$. Then if $dim F > dim E+ dim X$, $f$ can be extended to a monomorphismon on $X$ and any two such extensions are homotopic rel $Y$.
I completely do not follow the supplied proof.
It would be great if someone could provide references (or proof ) for each of the claims.
Consider the fibre bundle $Mon(E,F)$ where the fibers are the monomorphisms of $E_x rightarrow F_x$.
- This is homeomoprhic to $GL(n)/GL(n-m)$
This space is $n-m-1$ connected.
cross sections can be extended and are all homeomorphic if
$$ dim X le dim F- dim E -1 $$
algebraic-topology differential-topology vector-bundles fiber-bundles
algebraic-topology differential-topology vector-bundles fiber-bundles
asked Jan 31 at 16:55
CL.CL.
2,3102925
asked Jan 31 at 16:55
CL.CL.
2,3102925
asked Jan 31 at 16:55
CL.CL.
2,3102925
asked Jan 31 at 16:55
CL.CL.
2,3102925
asked Jan 31 at 16:55
CL.CL.
2,3102925
2,3102925
add a comment |
add a comment |
1 Answer
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$begingroup$
1) Let $V,W$ be $mathbb{K}$-vector spaces with $dim(V)leqdim (W)$. In particular there always there exists a linear monomorphism $varphi:Vhookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $Ain Gl(W)$ then $Avarphi:Vhookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)times Mon(V,W)rightarrow Mon(V,W),qquad (A,varphi)mapsto A varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $varphi_0in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(varphi_0)cong Mon(V,W),$$
where $Stab(varphi_0)leq Gl(W)$ is the stabiliser subgroup of $varphi_0$ with respect to the above action.
Now use the fixed $varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^perpleq W$ to get a direct sum decomposition $Voplus V^perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^perp)hookrightarrow Gl(W)$ which sends $B$ to $id_Voplus B$, and it is clear that
$$Stab(varphi_0)=Stab(V)cong Gl(V^perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^perp)cong Gl(Voplus V^perp)/Gl(V^perp)cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $Vcong mathbb{K}^n$ and $Wcongmathbb{K}^{n+m}$ and take $varphi_0$ to be the inclusion of $mathbb{K}^n$ in $mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)cong Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $xin X$, and so
$$Mon(E,F)cong bigcup_{xin X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}congmathbb{K}^n$ and $F_{x_0}congmathbb{K}^{n+m}$ for some $n,minmathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}cong Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+m})rightarrow Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(mathbb{K}^{n+m})$ is a Lie group, $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ is an open submanifold of $Mat_{ntimes(n+m)}(mathbb{K})$ and the action is smooth. Moreover $Gl(mathbb{K}^n)$ is closed in $Gl(mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(mathbb{K}^n,mathbb{K}^{n+1})$ is just the unit sphere
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1},$$
and you can see this directly by thinking of points of $S(mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+1})rightarrow S(mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1}=begin{cases}S^n&mathbb{K}=mathbb{R}\
S^{2n+1}&mathbb{K}=mathbb{C}\
S^{4n+3}&mathbb{K}=mathbb{H}end{cases}$$
so the map $Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(mathbb{K}^{n+1})rightarrow Gl(mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$. If $Erightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:Xrightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;pi_kF)$, the first living in $H^{n+1}(X;pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
$endgroup$
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
add a comment |
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$begingroup$
1) Let $V,W$ be $mathbb{K}$-vector spaces with $dim(V)leqdim (W)$. In particular there always there exists a linear monomorphism $varphi:Vhookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $Ain Gl(W)$ then $Avarphi:Vhookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)times Mon(V,W)rightarrow Mon(V,W),qquad (A,varphi)mapsto A varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $varphi_0in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(varphi_0)cong Mon(V,W),$$
where $Stab(varphi_0)leq Gl(W)$ is the stabiliser subgroup of $varphi_0$ with respect to the above action.
Now use the fixed $varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^perpleq W$ to get a direct sum decomposition $Voplus V^perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^perp)hookrightarrow Gl(W)$ which sends $B$ to $id_Voplus B$, and it is clear that
$$Stab(varphi_0)=Stab(V)cong Gl(V^perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^perp)cong Gl(Voplus V^perp)/Gl(V^perp)cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $Vcong mathbb{K}^n$ and $Wcongmathbb{K}^{n+m}$ and take $varphi_0$ to be the inclusion of $mathbb{K}^n$ in $mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)cong Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $xin X$, and so
$$Mon(E,F)cong bigcup_{xin X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}congmathbb{K}^n$ and $F_{x_0}congmathbb{K}^{n+m}$ for some $n,minmathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}cong Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+m})rightarrow Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(mathbb{K}^{n+m})$ is a Lie group, $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ is an open submanifold of $Mat_{ntimes(n+m)}(mathbb{K})$ and the action is smooth. Moreover $Gl(mathbb{K}^n)$ is closed in $Gl(mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(mathbb{K}^n,mathbb{K}^{n+1})$ is just the unit sphere
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1},$$
and you can see this directly by thinking of points of $S(mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+1})rightarrow S(mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1}=begin{cases}S^n&mathbb{K}=mathbb{R}\
S^{2n+1}&mathbb{K}=mathbb{C}\
S^{4n+3}&mathbb{K}=mathbb{H}end{cases}$$
so the map $Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(mathbb{K}^{n+1})rightarrow Gl(mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$. If $Erightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:Xrightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;pi_kF)$, the first living in $H^{n+1}(X;pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
$endgroup$
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
add a comment |
$begingroup$
1) Let $V,W$ be $mathbb{K}$-vector spaces with $dim(V)leqdim (W)$. In particular there always there exists a linear monomorphism $varphi:Vhookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $Ain Gl(W)$ then $Avarphi:Vhookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)times Mon(V,W)rightarrow Mon(V,W),qquad (A,varphi)mapsto A varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $varphi_0in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(varphi_0)cong Mon(V,W),$$
where $Stab(varphi_0)leq Gl(W)$ is the stabiliser subgroup of $varphi_0$ with respect to the above action.
Now use the fixed $varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^perpleq W$ to get a direct sum decomposition $Voplus V^perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^perp)hookrightarrow Gl(W)$ which sends $B$ to $id_Voplus B$, and it is clear that
$$Stab(varphi_0)=Stab(V)cong Gl(V^perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^perp)cong Gl(Voplus V^perp)/Gl(V^perp)cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $Vcong mathbb{K}^n$ and $Wcongmathbb{K}^{n+m}$ and take $varphi_0$ to be the inclusion of $mathbb{K}^n$ in $mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)cong Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $xin X$, and so
$$Mon(E,F)cong bigcup_{xin X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}congmathbb{K}^n$ and $F_{x_0}congmathbb{K}^{n+m}$ for some $n,minmathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}cong Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+m})rightarrow Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(mathbb{K}^{n+m})$ is a Lie group, $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ is an open submanifold of $Mat_{ntimes(n+m)}(mathbb{K})$ and the action is smooth. Moreover $Gl(mathbb{K}^n)$ is closed in $Gl(mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(mathbb{K}^n,mathbb{K}^{n+1})$ is just the unit sphere
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1},$$
and you can see this directly by thinking of points of $S(mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+1})rightarrow S(mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1}=begin{cases}S^n&mathbb{K}=mathbb{R}\
S^{2n+1}&mathbb{K}=mathbb{C}\
S^{4n+3}&mathbb{K}=mathbb{H}end{cases}$$
so the map $Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(mathbb{K}^{n+1})rightarrow Gl(mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$. If $Erightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:Xrightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;pi_kF)$, the first living in $H^{n+1}(X;pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
$endgroup$
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
add a comment |
$begingroup$
1) Let $V,W$ be $mathbb{K}$-vector spaces with $dim(V)leqdim (W)$. In particular there always there exists a linear monomorphism $varphi:Vhookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $Ain Gl(W)$ then $Avarphi:Vhookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)times Mon(V,W)rightarrow Mon(V,W),qquad (A,varphi)mapsto A varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $varphi_0in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(varphi_0)cong Mon(V,W),$$
where $Stab(varphi_0)leq Gl(W)$ is the stabiliser subgroup of $varphi_0$ with respect to the above action.
Now use the fixed $varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^perpleq W$ to get a direct sum decomposition $Voplus V^perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^perp)hookrightarrow Gl(W)$ which sends $B$ to $id_Voplus B$, and it is clear that
$$Stab(varphi_0)=Stab(V)cong Gl(V^perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^perp)cong Gl(Voplus V^perp)/Gl(V^perp)cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $Vcong mathbb{K}^n$ and $Wcongmathbb{K}^{n+m}$ and take $varphi_0$ to be the inclusion of $mathbb{K}^n$ in $mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)cong Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $xin X$, and so
$$Mon(E,F)cong bigcup_{xin X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}congmathbb{K}^n$ and $F_{x_0}congmathbb{K}^{n+m}$ for some $n,minmathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}cong Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+m})rightarrow Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(mathbb{K}^{n+m})$ is a Lie group, $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ is an open submanifold of $Mat_{ntimes(n+m)}(mathbb{K})$ and the action is smooth. Moreover $Gl(mathbb{K}^n)$ is closed in $Gl(mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(mathbb{K}^n,mathbb{K}^{n+1})$ is just the unit sphere
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1},$$
and you can see this directly by thinking of points of $S(mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+1})rightarrow S(mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1}=begin{cases}S^n&mathbb{K}=mathbb{R}\
S^{2n+1}&mathbb{K}=mathbb{C}\
S^{4n+3}&mathbb{K}=mathbb{H}end{cases}$$
so the map $Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(mathbb{K}^{n+1})rightarrow Gl(mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$. If $Erightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:Xrightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;pi_kF)$, the first living in $H^{n+1}(X;pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
$endgroup$
1) Let $V,W$ be $mathbb{K}$-vector spaces with $dim(V)leqdim (W)$. In particular there always there exists a linear monomorphism $varphi:Vhookrightarrow W$, a fact you can verify easily by choosing bases for $V$ and $W$ (a vector space always admits a basis). Now if $Ain Gl(W)$ then $Avarphi:Vhookrightarrow W$ is also monic, and this observation gives us an action
$$Gl(W)times Mon(V,W)rightarrow Mon(V,W),qquad (A,varphi)mapsto A varphi.$$
It's not difficult to check that this action is transitive, again by choosing bases. Moreover all the spaces and maps are suitably nice, so if we fix a basepoint monomorphism $varphi_0in Mon(V,W)$, then we get an induced homeomorphism
$$Gl(W)/Stab(varphi_0)cong Mon(V,W),$$
where $Stab(varphi_0)leq Gl(W)$ is the stabiliser subgroup of $varphi_0$ with respect to the above action.
Now use the fixed $varphi_0$ to identify $V$ as a subspace of $W$ and choose a complement $V^perpleq W$ to get a direct sum decomposition $Voplus V^perp$. Vector subspaces always admit complements - just choose bases. A slightly fancier way to generate a complement (as the notation suggets) is to fix an arbitrary inner product on $W$ (choose a basis) and let $V^perp$ be the orthogonal complement with respect to this inner product.
Now there is a subgroup inclusion $Gl(V^perp)hookrightarrow Gl(W)$ which sends $B$ to $id_Voplus B$, and it is clear that
$$Stab(varphi_0)=Stab(V)cong Gl(V^perp)$$
with respect to this inclusion. Putting everything together we have a homeomorphism
$$Gl(W)/Gl(V^perp)cong Gl(Voplus V^perp)/Gl(V^perp)cong Mon(V,W).$$
To see things most clearly fix bases for $V$, $W$ (sigh) so that $Vcong mathbb{K}^n$ and $Wcongmathbb{K}^{n+m}$ and take $varphi_0$ to be the inclusion of $mathbb{K}^n$ in $mathbb{K}^{n+m}$ as the first $n$ non-zero coordinates. In this case our previous homeomorphism is just
$$Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)cong Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
Now this is all unparametrised, but the same construction can be carried out fibrewise. With your notation we have $Mon(E,F)_x=Mon(E_x,F_x)$ for $xin X$, and so
$$Mon(E,F)cong bigcup_{xin X}Mon(E_x,F_x).$$
The fibre over a fixed basepoint $x_0in X$ is just $Mon(E,F)_{x_0}=Mon(E_{x_0},F_{x_0})$, and since $E_{x_0}congmathbb{K}^n$ and $F_{x_0}congmathbb{K}^{n+m}$ for some $n,minmathbb{N}_0$ by assumption we have
$$Mon(E,F)_{x_0}cong Gl(mathbb{K}^{n+m})/Gl(mathbb{K}^m)$$
as above.
2) What we've actually shown above is that we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+m})rightarrow Mon(mathbb{K}^n,mathbb{K}^{n+m}).$$
You can get this by running through the standard theorems. $Gl(mathbb{K}^{n+m})$ is a Lie group, $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ is an open submanifold of $Mat_{ntimes(n+m)}(mathbb{K})$ and the action is smooth. Moreover $Gl(mathbb{K}^n)$ is closed in $Gl(mathbb{K}^{n+m})$, so the projection onto the oribt space is the just projection of a Lie group onto its quotient by a closed subgroup (in particular it admits local sections and is thus a fibration).
Now take the fibration sequence above and set $m=1$. Then $Mon(mathbb{K}^n,mathbb{K}^{n+1})$ is just the unit sphere
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1},$$
and you can see this directly by thinking of points of $S(mathbb{K}^{n+1})$ as $1$-dimensional subspaces of $mathbb{K}^{n+1}$. Thus we have a fibration sequence
$$Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^{n+1})rightarrow S(mathbb{K}^{n+1})$$
and in particular a long exact sequence of homotopy groups. We have
$$S(mathbb{K}^{n+1})={xinmathbb{K}^{n+1}mid |x|^2=1}=begin{cases}S^n&mathbb{K}=mathbb{R}\
S^{2n+1}&mathbb{K}=mathbb{C}\
S^{4n+3}&mathbb{K}=mathbb{H}end{cases}$$
so the map $Gl(mathbb{K}^n)rightarrow Gl(mathbb{K}^n)$ is as connected as the indicated sphere is. In the real case this map is $(n-1)$-connected. On the other hand, the map$Gl(mathbb{K}^{n+1})rightarrow Gl(mathbb{K}^{n+1})$ is $(n+1)$-connected, and therefore the composite inclusion $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+2})$ is as connected as the fist inclusion, so is $(n-1)$-connected.
Iterating this gives us that $Gl(mathbb{K}^n)hookrightarrow Gl(mathbb{K}^{n+m})$ is $(n-1)$-connected, so from our fibration sequence the first non-trivial homotopy group of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$ occurs in degree $n$, so the space is $(n-1)$-connected. This matches up with your statement when you recall that they have written $n$ where I have written $n+m$, and $n-m$ where I have written $n$.
3) Your final query is just an exercise in obstruction theory using what we now now about the connectivity of $Mon(mathbb{K}^n,mathbb{K}^{n+m})$. If $Erightarrow B$ is a suitable fibration with $(n-1)$-connected fibre, and you are given a map $f:Xrightarrow B$, then there are a chain of obstructions to lifting to map into $E$ which covers $f$, and these lie in the group $H^{k+1}(X;pi_kF)$, the first living in $H^{n+1}(X;pi_nF)$. If all these obstructions vanish then a lift can be found. Clearly if $dim(X)<n$, then all the obstructions live in trivial groups, so vanish showing that a lift exists.
I'm afraid that here is not the best place to explain much further details, but a good place to strat reading about obstruction theory is in Hatcher's book Algebraic Toplogy on pg. 415. Davis and Kirk also do a very good treatement in their book (whose title escapes me right now). Classic references (although a little more difficult) are Steenrod, Whitehead and Spanier.
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
answered Feb 1 at 11:36
TyroneTyrone
5,31711226
5,31711226
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
add a comment |
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
$begingroup$
The book title is "Lecture notes in algebraic topology". I agree this is one of the best places to first learn obstruction theory.
$endgroup$
– user98602
Feb 1 at 14:11
add a comment |
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