Lefschetz number of constant map
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Consider a map from a connected $n$-manifold $M$ to itself defined by $f(p) = c$ for some fixed $c in M$. This is a Lefschetz map since the subset $M times {c }$ intersects the diagonal $Delta$ transversely.
We can compute the Lefchetz number in two different ways, first algerbaic-topologically as the alternating sum of the trace of $f_*$ on $H_*(M)$. Then $f_*$ is the identity on $H_0(M)$ and zero on the higher homology groups. So the trace of $f_*$ on $H_0(M)$ is $1$ and zero on the higher homology groups. Therefore we get that the Lefschetz number of $f$ is 1 using the algerbaic-topological method, let's write $L^{AT}(f) = 1$.
Now consider the differential topological description given by the Lefschetz-Hopf theorem as the sum of the local Lefschetz number of the fixed points. Clearly $f$ has precisely one fixed point $c$. The Local Lefschetz number $L_c (f)$ is equal to the sign of the determinant of $df_c - Id$. However, here we have $df_c = 0$ since $f$ is the constant map. So we have $det(df_c - Id) = det(0-Id) = det(-Id) = (-1)^n$ and so we have $L_c(f) = (-1)^n$ where $n$ is the dimension of the manifold. In particular this allows us to calculate the Lefschetz number differential-topologically and get $(-1)^n$ instead, let's write $L^{DT}(f) = (-1)^n$.
Now observe for odd dimensional manifolds we get $L^{DT}(f) = -1$ while $L^{AT}(f)=1$. However, the Lefschetz-Hopf theorem says these two definitions of the Lefschetz number should agree.
Question: Why do we have $L^{AT}(f) neq L^{DT} (f)$ for odd dimensional manifolds? Is there some error in my calculations?
algebraic-topology differential-topology fixed-point-theorems
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
$endgroup$
add a comment |
$begingroup$
Consider a map from a connected $n$-manifold $M$ to itself defined by $f(p) = c$ for some fixed $c in M$. This is a Lefschetz map since the subset $M times {c }$ intersects the diagonal $Delta$ transversely.
We can compute the Lefchetz number in two different ways, first algerbaic-topologically as the alternating sum of the trace of $f_*$ on $H_*(M)$. Then $f_*$ is the identity on $H_0(M)$ and zero on the higher homology groups. So the trace of $f_*$ on $H_0(M)$ is $1$ and zero on the higher homology groups. Therefore we get that the Lefschetz number of $f$ is 1 using the algerbaic-topological method, let's write $L^{AT}(f) = 1$.
Now consider the differential topological description given by the Lefschetz-Hopf theorem as the sum of the local Lefschetz number of the fixed points. Clearly $f$ has precisely one fixed point $c$. The Local Lefschetz number $L_c (f)$ is equal to the sign of the determinant of $df_c - Id$. However, here we have $df_c = 0$ since $f$ is the constant map. So we have $det(df_c - Id) = det(0-Id) = det(-Id) = (-1)^n$ and so we have $L_c(f) = (-1)^n$ where $n$ is the dimension of the manifold. In particular this allows us to calculate the Lefschetz number differential-topologically and get $(-1)^n$ instead, let's write $L^{DT}(f) = (-1)^n$.
Now observe for odd dimensional manifolds we get $L^{DT}(f) = -1$ while $L^{AT}(f)=1$. However, the Lefschetz-Hopf theorem says these two definitions of the Lefschetz number should agree.
Question: Why do we have $L^{AT}(f) neq L^{DT} (f)$ for odd dimensional manifolds? Is there some error in my calculations?
algebraic-topology differential-topology fixed-point-theorems
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
$endgroup$
1
$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21
add a comment |
$begingroup$
Consider a map from a connected $n$-manifold $M$ to itself defined by $f(p) = c$ for some fixed $c in M$. This is a Lefschetz map since the subset $M times {c }$ intersects the diagonal $Delta$ transversely.
We can compute the Lefchetz number in two different ways, first algerbaic-topologically as the alternating sum of the trace of $f_*$ on $H_*(M)$. Then $f_*$ is the identity on $H_0(M)$ and zero on the higher homology groups. So the trace of $f_*$ on $H_0(M)$ is $1$ and zero on the higher homology groups. Therefore we get that the Lefschetz number of $f$ is 1 using the algerbaic-topological method, let's write $L^{AT}(f) = 1$.
Now consider the differential topological description given by the Lefschetz-Hopf theorem as the sum of the local Lefschetz number of the fixed points. Clearly $f$ has precisely one fixed point $c$. The Local Lefschetz number $L_c (f)$ is equal to the sign of the determinant of $df_c - Id$. However, here we have $df_c = 0$ since $f$ is the constant map. So we have $det(df_c - Id) = det(0-Id) = det(-Id) = (-1)^n$ and so we have $L_c(f) = (-1)^n$ where $n$ is the dimension of the manifold. In particular this allows us to calculate the Lefschetz number differential-topologically and get $(-1)^n$ instead, let's write $L^{DT}(f) = (-1)^n$.
Now observe for odd dimensional manifolds we get $L^{DT}(f) = -1$ while $L^{AT}(f)=1$. However, the Lefschetz-Hopf theorem says these two definitions of the Lefschetz number should agree.
Question: Why do we have $L^{AT}(f) neq L^{DT} (f)$ for odd dimensional manifolds? Is there some error in my calculations?
algebraic-topology differential-topology fixed-point-theorems
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
$endgroup$
Consider a map from a connected $n$-manifold $M$ to itself defined by $f(p) = c$ for some fixed $c in M$. This is a Lefschetz map since the subset $M times {c }$ intersects the diagonal $Delta$ transversely.
We can compute the Lefchetz number in two different ways, first algerbaic-topologically as the alternating sum of the trace of $f_*$ on $H_*(M)$. Then $f_*$ is the identity on $H_0(M)$ and zero on the higher homology groups. So the trace of $f_*$ on $H_0(M)$ is $1$ and zero on the higher homology groups. Therefore we get that the Lefschetz number of $f$ is 1 using the algerbaic-topological method, let's write $L^{AT}(f) = 1$.
Now consider the differential topological description given by the Lefschetz-Hopf theorem as the sum of the local Lefschetz number of the fixed points. Clearly $f$ has precisely one fixed point $c$. The Local Lefschetz number $L_c (f)$ is equal to the sign of the determinant of $df_c - Id$. However, here we have $df_c = 0$ since $f$ is the constant map. So we have $det(df_c - Id) = det(0-Id) = det(-Id) = (-1)^n$ and so we have $L_c(f) = (-1)^n$ where $n$ is the dimension of the manifold. In particular this allows us to calculate the Lefschetz number differential-topologically and get $(-1)^n$ instead, let's write $L^{DT}(f) = (-1)^n$.
Now observe for odd dimensional manifolds we get $L^{DT}(f) = -1$ while $L^{AT}(f)=1$. However, the Lefschetz-Hopf theorem says these two definitions of the Lefschetz number should agree.
Question: Why do we have $L^{AT}(f) neq L^{DT} (f)$ for odd dimensional manifolds? Is there some error in my calculations?
algebraic-topology differential-topology fixed-point-theorems
algebraic-topology differential-topology fixed-point-theorems
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
asked Jan 16 at 20:08
Kai NakamuraKai Nakamura
31719
31719
1
$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21
add a comment |
1
$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21
1
1
$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21
$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21
add a comment |
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$begingroup$
Whatever author you're reading is just using a convention that doesn't agree with the other. This is harmless. Your formula just then becomes $L^{AT}(f) = (-1)^n L^{DT}(f)$, so they differ by a uniform sign.
$endgroup$
– user98602
Jan 16 at 21:21