Mixing
Exercise about retract of $GL(n, mathbb{C})$ homeomorphic to $S^1$
$begingroup$
I have been working through the following exercise:
Consider $X = GL (n, mathbb{C})$ with respect to the Euclidean topology in $mathbb{C}^{n^2}$.
Find a subspace $Y subset X$ such that $Y$ is homemorphic to $S^1$ and $Y$ is a retract of $X$.
Show that the fundamental group $pi_1 (X, x)$ is infinite for all $x in Y$.
Let's consider the set $Y$ which consists of all matrices
begin{equation*}
begin{pmatrix}
z & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix},
end{equation*}
with $z in mathbb{C}$ such that $|z| = 1$.
Then $Y$ is a subspace of $X$ and it is homeomorphic to $S^1$.
Let $r : X rightarrow Y$ be the map defined by
begin{equation*}
r(A) =
begin{pmatrix}
frac{det A}{|det A|} & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix}
end{equation*}
for all matrices $A in X$. This map is continuous because the determinant is a continuous map and every matrix in $X$ has non-zero determinant. Besides, $r$ is a retraction of $X$ to $Y$, because $r|_Y$ is the identity map on $Y$.
Let's consider the inclusion $i : Y rightarrow X$ and a point $x in Y$. Since $r$ is a retraction, the group homomorphism
begin{equation*}
i_* : pi_1 (Y, x) rightarrow pi_1(X, x)
end{equation*}
is injective. Then $pi_1(X, x)$ is infinite, because $pi_1 (Y, x)$ is an infinite (cyclic) group.
Is this solution acceptable?
proof-verification algebraic-topology
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
$endgroup$
add a comment |
$begingroup$
I have been working through the following exercise:
Consider $X = GL (n, mathbb{C})$ with respect to the Euclidean topology in $mathbb{C}^{n^2}$.
Find a subspace $Y subset X$ such that $Y$ is homemorphic to $S^1$ and $Y$ is a retract of $X$.
Show that the fundamental group $pi_1 (X, x)$ is infinite for all $x in Y$.
Let's consider the set $Y$ which consists of all matrices
begin{equation*}
begin{pmatrix}
z & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix},
end{equation*}
with $z in mathbb{C}$ such that $|z| = 1$.
Then $Y$ is a subspace of $X$ and it is homeomorphic to $S^1$.
Let $r : X rightarrow Y$ be the map defined by
begin{equation*}
r(A) =
begin{pmatrix}
frac{det A}{|det A|} & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix}
end{equation*}
for all matrices $A in X$. This map is continuous because the determinant is a continuous map and every matrix in $X$ has non-zero determinant. Besides, $r$ is a retraction of $X$ to $Y$, because $r|_Y$ is the identity map on $Y$.
Let's consider the inclusion $i : Y rightarrow X$ and a point $x in Y$. Since $r$ is a retraction, the group homomorphism
begin{equation*}
i_* : pi_1 (Y, x) rightarrow pi_1(X, x)
end{equation*}
is injective. Then $pi_1(X, x)$ is infinite, because $pi_1 (Y, x)$ is an infinite (cyclic) group.
Is this solution acceptable?
proof-verification algebraic-topology
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
$endgroup$
3
$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41
add a comment |
$begingroup$
I have been working through the following exercise:
Consider $X = GL (n, mathbb{C})$ with respect to the Euclidean topology in $mathbb{C}^{n^2}$.
Find a subspace $Y subset X$ such that $Y$ is homemorphic to $S^1$ and $Y$ is a retract of $X$.
Show that the fundamental group $pi_1 (X, x)$ is infinite for all $x in Y$.
Let's consider the set $Y$ which consists of all matrices
begin{equation*}
begin{pmatrix}
z & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix},
end{equation*}
with $z in mathbb{C}$ such that $|z| = 1$.
Then $Y$ is a subspace of $X$ and it is homeomorphic to $S^1$.
Let $r : X rightarrow Y$ be the map defined by
begin{equation*}
r(A) =
begin{pmatrix}
frac{det A}{|det A|} & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix}
end{equation*}
for all matrices $A in X$. This map is continuous because the determinant is a continuous map and every matrix in $X$ has non-zero determinant. Besides, $r$ is a retraction of $X$ to $Y$, because $r|_Y$ is the identity map on $Y$.
Let's consider the inclusion $i : Y rightarrow X$ and a point $x in Y$. Since $r$ is a retraction, the group homomorphism
begin{equation*}
i_* : pi_1 (Y, x) rightarrow pi_1(X, x)
end{equation*}
is injective. Then $pi_1(X, x)$ is infinite, because $pi_1 (Y, x)$ is an infinite (cyclic) group.
Is this solution acceptable?
proof-verification algebraic-topology
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
$endgroup$
I have been working through the following exercise:
Consider $X = GL (n, mathbb{C})$ with respect to the Euclidean topology in $mathbb{C}^{n^2}$.
Find a subspace $Y subset X$ such that $Y$ is homemorphic to $S^1$ and $Y$ is a retract of $X$.
Show that the fundamental group $pi_1 (X, x)$ is infinite for all $x in Y$.
Let's consider the set $Y$ which consists of all matrices
begin{equation*}
begin{pmatrix}
z & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix},
end{equation*}
with $z in mathbb{C}$ such that $|z| = 1$.
Then $Y$ is a subspace of $X$ and it is homeomorphic to $S^1$.
Let $r : X rightarrow Y$ be the map defined by
begin{equation*}
r(A) =
begin{pmatrix}
frac{det A}{|det A|} & 0 & ldots & 0 \
0 & 1 & ldots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & ldots & 1
end{pmatrix}
end{equation*}
for all matrices $A in X$. This map is continuous because the determinant is a continuous map and every matrix in $X$ has non-zero determinant. Besides, $r$ is a retraction of $X$ to $Y$, because $r|_Y$ is the identity map on $Y$.
Let's consider the inclusion $i : Y rightarrow X$ and a point $x in Y$. Since $r$ is a retraction, the group homomorphism
begin{equation*}
i_* : pi_1 (Y, x) rightarrow pi_1(X, x)
end{equation*}
is injective. Then $pi_1(X, x)$ is infinite, because $pi_1 (Y, x)$ is an infinite (cyclic) group.
Is this solution acceptable?
proof-verification algebraic-topology
proof-verification algebraic-topology
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
edited Jan 28 at 21:58
Paolo
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
asked Aug 28 '18 at 21:32
PaoloPaolo
6371420
6371420
3
$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41
add a comment |
3
$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41
3
3
$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41
add a comment |
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$begingroup$
This looks fine. For extra credit, can you show $GL(n,Bbb C)$ deformation retracts to $U(n)$? :P
$endgroup$
– Ted Shifrin
Aug 28 '18 at 21:35
$begingroup$
@TedShifrin I'll think about it :)
$endgroup$
– Paolo
Aug 28 '18 at 21:41