Paracompactness of the projectified bundle over a paracompact space
$begingroup$
Consider a complex rank $n$ Vector Bundle $V rightarrow X$. It is a standard Argument in the construction of Chern classes to consider the orthogonal complement to the tautological line bundle carried by $P(V)$, where $P(V)$ is the projectified bundle over $X$ with fiber $mathbb C P^{n-1}$, basically $V setminus X$ with $mathbb C$-multiplication factored out.
I have found numerous references to the fact that $P(V)$ is paracompact, provided that $X$ is, but no rigorous proof. The idea must somehow be to use that products with one factor paracompact and the other compact are paracompact, which would take care of the case $V cong X times mathbb C^n$. As to other cases, say, $X= U_1 cup U_2$ and $V$ is trivial over the $U_i$, my problem would be that the $U_i$ need not be paracompact, as they are open in $X$. Perhaps there is a way around this by looking at gluing constructions to recover $V$ as $(U_1 times mathbb C^n) dotcup (U_2 times mathbb C^n)/ sim$, but I am stuck. Any pointers?
general-topology vector-bundles paracompactness
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
add a comment |
$begingroup$
Consider a complex rank $n$ Vector Bundle $V rightarrow X$. It is a standard Argument in the construction of Chern classes to consider the orthogonal complement to the tautological line bundle carried by $P(V)$, where $P(V)$ is the projectified bundle over $X$ with fiber $mathbb C P^{n-1}$, basically $V setminus X$ with $mathbb C$-multiplication factored out.
I have found numerous references to the fact that $P(V)$ is paracompact, provided that $X$ is, but no rigorous proof. The idea must somehow be to use that products with one factor paracompact and the other compact are paracompact, which would take care of the case $V cong X times mathbb C^n$. As to other cases, say, $X= U_1 cup U_2$ and $V$ is trivial over the $U_i$, my problem would be that the $U_i$ need not be paracompact, as they are open in $X$. Perhaps there is a way around this by looking at gluing constructions to recover $V$ as $(U_1 times mathbb C^n) dotcup (U_2 times mathbb C^n)/ sim$, but I am stuck. Any pointers?
general-topology vector-bundles paracompactness
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
add a comment |
$begingroup$
Consider a complex rank $n$ Vector Bundle $V rightarrow X$. It is a standard Argument in the construction of Chern classes to consider the orthogonal complement to the tautological line bundle carried by $P(V)$, where $P(V)$ is the projectified bundle over $X$ with fiber $mathbb C P^{n-1}$, basically $V setminus X$ with $mathbb C$-multiplication factored out.
I have found numerous references to the fact that $P(V)$ is paracompact, provided that $X$ is, but no rigorous proof. The idea must somehow be to use that products with one factor paracompact and the other compact are paracompact, which would take care of the case $V cong X times mathbb C^n$. As to other cases, say, $X= U_1 cup U_2$ and $V$ is trivial over the $U_i$, my problem would be that the $U_i$ need not be paracompact, as they are open in $X$. Perhaps there is a way around this by looking at gluing constructions to recover $V$ as $(U_1 times mathbb C^n) dotcup (U_2 times mathbb C^n)/ sim$, but I am stuck. Any pointers?
general-topology vector-bundles paracompactness
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
Consider a complex rank $n$ Vector Bundle $V rightarrow X$. It is a standard Argument in the construction of Chern classes to consider the orthogonal complement to the tautological line bundle carried by $P(V)$, where $P(V)$ is the projectified bundle over $X$ with fiber $mathbb C P^{n-1}$, basically $V setminus X$ with $mathbb C$-multiplication factored out.
I have found numerous references to the fact that $P(V)$ is paracompact, provided that $X$ is, but no rigorous proof. The idea must somehow be to use that products with one factor paracompact and the other compact are paracompact, which would take care of the case $V cong X times mathbb C^n$. As to other cases, say, $X= U_1 cup U_2$ and $V$ is trivial over the $U_i$, my problem would be that the $U_i$ need not be paracompact, as they are open in $X$. Perhaps there is a way around this by looking at gluing constructions to recover $V$ as $(U_1 times mathbb C^n) dotcup (U_2 times mathbb C^n)/ sim$, but I am stuck. Any pointers?
general-topology vector-bundles paracompactness
general-topology vector-bundles paracompactness
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
edited Dec 9 '18 at 3:28
Eric Wofsey
184k13212338
184k13212338
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
asked Jan 11 '17 at 22:57
TheHumanHighwayTheHumanHighway
711113
711113
add a comment |
add a comment |
1 Answer
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I think I got it. The idea is to take an open cover ${U_alpha}_{alpha in J}$ and first prove a "Shrinking" lemma for paracompact spaces. Even though $X$ is normal, given its paracompactness, the standard Shrinking lemma (for a proof see Munkres) needs a finite cover (I guess the proof goes through for a countably infinite one as well). But we can find a locally finite refinement with same index set, ${V_alpha}_{alpha in J}$ such that it still is a cover and $V_alpha subset bar{V}_alphasubset U_alpha$. Now if $V rightarrow X$ is trivial over each $U_alpha$, then certainly over each $bar V_alpha$, too.
But closed subsets of paracompact spaces are paracompact, and since $P(V)|_{bar V_alpha} cong bar V_alpha times mathbb CP^{n-1}$, the disjoint union $amalg_alpha (bar V_alpha times mathbb CP^{n-1})$ is paracompact, for products consisting of a paracompact and a compact space are paracompact, and disjoint unions of paracompact spaces are paracompact. Now the projections induced from the gluing construction from which we recover $P(V) rightarrow X$ are closed, hence by Michael's theorem $P(V) cong amalg_alpha (bar V_alpha times mathbb CP^{n-1})/sim$ is paracompact.
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
add a comment |
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$begingroup$
I think I got it. The idea is to take an open cover ${U_alpha}_{alpha in J}$ and first prove a "Shrinking" lemma for paracompact spaces. Even though $X$ is normal, given its paracompactness, the standard Shrinking lemma (for a proof see Munkres) needs a finite cover (I guess the proof goes through for a countably infinite one as well). But we can find a locally finite refinement with same index set, ${V_alpha}_{alpha in J}$ such that it still is a cover and $V_alpha subset bar{V}_alphasubset U_alpha$. Now if $V rightarrow X$ is trivial over each $U_alpha$, then certainly over each $bar V_alpha$, too.
But closed subsets of paracompact spaces are paracompact, and since $P(V)|_{bar V_alpha} cong bar V_alpha times mathbb CP^{n-1}$, the disjoint union $amalg_alpha (bar V_alpha times mathbb CP^{n-1})$ is paracompact, for products consisting of a paracompact and a compact space are paracompact, and disjoint unions of paracompact spaces are paracompact. Now the projections induced from the gluing construction from which we recover $P(V) rightarrow X$ are closed, hence by Michael's theorem $P(V) cong amalg_alpha (bar V_alpha times mathbb CP^{n-1})/sim$ is paracompact.
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
add a comment |
$begingroup$
I think I got it. The idea is to take an open cover ${U_alpha}_{alpha in J}$ and first prove a "Shrinking" lemma for paracompact spaces. Even though $X$ is normal, given its paracompactness, the standard Shrinking lemma (for a proof see Munkres) needs a finite cover (I guess the proof goes through for a countably infinite one as well). But we can find a locally finite refinement with same index set, ${V_alpha}_{alpha in J}$ such that it still is a cover and $V_alpha subset bar{V}_alphasubset U_alpha$. Now if $V rightarrow X$ is trivial over each $U_alpha$, then certainly over each $bar V_alpha$, too.
But closed subsets of paracompact spaces are paracompact, and since $P(V)|_{bar V_alpha} cong bar V_alpha times mathbb CP^{n-1}$, the disjoint union $amalg_alpha (bar V_alpha times mathbb CP^{n-1})$ is paracompact, for products consisting of a paracompact and a compact space are paracompact, and disjoint unions of paracompact spaces are paracompact. Now the projections induced from the gluing construction from which we recover $P(V) rightarrow X$ are closed, hence by Michael's theorem $P(V) cong amalg_alpha (bar V_alpha times mathbb CP^{n-1})/sim$ is paracompact.
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
add a comment |
$begingroup$
I think I got it. The idea is to take an open cover ${U_alpha}_{alpha in J}$ and first prove a "Shrinking" lemma for paracompact spaces. Even though $X$ is normal, given its paracompactness, the standard Shrinking lemma (for a proof see Munkres) needs a finite cover (I guess the proof goes through for a countably infinite one as well). But we can find a locally finite refinement with same index set, ${V_alpha}_{alpha in J}$ such that it still is a cover and $V_alpha subset bar{V}_alphasubset U_alpha$. Now if $V rightarrow X$ is trivial over each $U_alpha$, then certainly over each $bar V_alpha$, too.
But closed subsets of paracompact spaces are paracompact, and since $P(V)|_{bar V_alpha} cong bar V_alpha times mathbb CP^{n-1}$, the disjoint union $amalg_alpha (bar V_alpha times mathbb CP^{n-1})$ is paracompact, for products consisting of a paracompact and a compact space are paracompact, and disjoint unions of paracompact spaces are paracompact. Now the projections induced from the gluing construction from which we recover $P(V) rightarrow X$ are closed, hence by Michael's theorem $P(V) cong amalg_alpha (bar V_alpha times mathbb CP^{n-1})/sim$ is paracompact.
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
$endgroup$
I think I got it. The idea is to take an open cover ${U_alpha}_{alpha in J}$ and first prove a "Shrinking" lemma for paracompact spaces. Even though $X$ is normal, given its paracompactness, the standard Shrinking lemma (for a proof see Munkres) needs a finite cover (I guess the proof goes through for a countably infinite one as well). But we can find a locally finite refinement with same index set, ${V_alpha}_{alpha in J}$ such that it still is a cover and $V_alpha subset bar{V}_alphasubset U_alpha$. Now if $V rightarrow X$ is trivial over each $U_alpha$, then certainly over each $bar V_alpha$, too.
But closed subsets of paracompact spaces are paracompact, and since $P(V)|_{bar V_alpha} cong bar V_alpha times mathbb CP^{n-1}$, the disjoint union $amalg_alpha (bar V_alpha times mathbb CP^{n-1})$ is paracompact, for products consisting of a paracompact and a compact space are paracompact, and disjoint unions of paracompact spaces are paracompact. Now the projections induced from the gluing construction from which we recover $P(V) rightarrow X$ are closed, hence by Michael's theorem $P(V) cong amalg_alpha (bar V_alpha times mathbb CP^{n-1})/sim$ is paracompact.
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
answered Feb 1 '17 at 9:57
TheHumanHighwayTheHumanHighway
711113
711113
add a comment |
add a comment |
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