Lefschetz number of constant map












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$begingroup$


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?










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$endgroup$








  • 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


















1












$begingroup$


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?










share|cite|improve this question









$endgroup$








  • 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
















1












1








1





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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asked Jan 16 at 20:08









Kai NakamuraKai Nakamura

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  • 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




    $begingroup$
    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












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