$K$-Theory of operators I, Higson notes Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Special elements in the $C^*$ algebra $A otimes mathcalK$.K-theory for non-separable C*-algebrasK-theory, $K_0$ of algebra of compact operatorsWhat is the motivation of studying $P[A]$ in operator K-theory?K-theory of $C_0(X)$Periodicity of Fredholm operators for proving Bott periodicityIn what way is $sigma(L) in K(T^*X)$ in the $K$-theoretic formulation of the Atiyah-Singer index theorem?Operator K-theory and Topological K-theoryInduced $K$-theory maps between $C^*$ algebras.Understanding the map from $K_0(A)$ to homotopy class of maps,Special elements in the $C^*$ algebra $A otimes mathcalK$.
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$K$-Theory of operators I, Higson notes
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Special elements in the $C^*$ algebra $A otimes mathcalK$.K-theory for non-separable C*-algebrasK-theory, $K_0$ of algebra of compact operatorsWhat is the motivation of studying $P[A]$ in operator K-theory?K-theory of $C_0(X)$Periodicity of Fredholm operators for proving Bott periodicityIn what way is $sigma(L) in K(T^*X)$ in the $K$-theoretic formulation of the Atiyah-Singer index theorem?Operator K-theory and Topological K-theoryInduced $K$-theory maps between $C^*$ algebras.Understanding the map from $K_0(A)$ to homotopy class of maps,Special elements in the $C^*$ algebra $A otimes mathcalK$.
$begingroup$
I am having trouble understanding the following statement:
3.20 Proposition, pg44: Let $D$ be a symmetric, odd graded elliptic operator on a graded vector bundle $S$ over a compact manifold. The element $[phi_D] in K(Bbb C) cong Bbb Z$ is the Fredholm index of the operator $D$.
What exactly are used in making sense of this statement:
Def 2.10, page 19. A Graded $*$-homomoprhism $phi_D:mathcalS rightarrow mathcalK(H)$ from spectral theorem.
Prop 3.17. There is an isomoprhism $$Phi:K(A) rightarrow [mathcalS, A otimes mathcalK(H)]$$
What the above two means is explain in my other post.
The proof goes as follows:
We have a homotopy of $*$-homomoprhisms $phi_s^-1D(f) = f(s^-1D)$. At $s=1$ we have $phi_D$ at $s=0$ we have the homomorphism of $f mapsto f(0)P$ where $P$ is projection onto the kernel of $D$.
How does one obtain continuity at $s=0$ - why is the resulting map as described?
This corresponds to the integer $dim(ker D cap H_+) - dim (ker D cap H_-)$.
How does one make this computation?
operator-algebras k-theory topological-k-theory
asked Mar 18 at 17:55
CL.CL.
2,2723925
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding the following statement:
3.20 Proposition, pg44: Let $D$ be a symmetric, odd graded elliptic operator on a graded vector bundle $S$ over a compact manifold. The element $[phi_D] in K(Bbb C) cong Bbb Z$ is the Fredholm index of the operator $D$.
What exactly are used in making sense of this statement:
Def 2.10, page 19. A Graded $*$-homomoprhism $phi_D:mathcalS rightarrow mathcalK(H)$ from spectral theorem.
Prop 3.17. There is an isomoprhism $$Phi:K(A) rightarrow [mathcalS, A otimes mathcalK(H)]$$
What the above two means is explain in my other post.
The proof goes as follows:
We have a homotopy of $*$-homomoprhisms $phi_s^-1D(f) = f(s^-1D)$. At $s=1$ we have $phi_D$ at $s=0$ we have the homomorphism of $f mapsto f(0)P$ where $P$ is projection onto the kernel of $D$.
How does one obtain continuity at $s=0$ - why is the resulting map as described?
This corresponds to the integer $dim(ker D cap H_+) - dim (ker D cap H_-)$.
How does one make this computation?
operator-algebras k-theory topological-k-theory
asked Mar 18 at 17:55
CL.CL.
2,2723925
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding the following statement:
3.20 Proposition, pg44: Let $D$ be a symmetric, odd graded elliptic operator on a graded vector bundle $S$ over a compact manifold. The element $[phi_D] in K(Bbb C) cong Bbb Z$ is the Fredholm index of the operator $D$.
What exactly are used in making sense of this statement:
Def 2.10, page 19. A Graded $*$-homomoprhism $phi_D:mathcalS rightarrow mathcalK(H)$ from spectral theorem.
Prop 3.17. There is an isomoprhism $$Phi:K(A) rightarrow [mathcalS, A otimes mathcalK(H)]$$
What the above two means is explain in my other post.
The proof goes as follows:
We have a homotopy of $*$-homomoprhisms $phi_s^-1D(f) = f(s^-1D)$. At $s=1$ we have $phi_D$ at $s=0$ we have the homomorphism of $f mapsto f(0)P$ where $P$ is projection onto the kernel of $D$.
How does one obtain continuity at $s=0$ - why is the resulting map as described?
This corresponds to the integer $dim(ker D cap H_+) - dim (ker D cap H_-)$.
How does one make this computation?
operator-algebras k-theory topological-k-theory
asked Mar 18 at 17:55
CL.CL.
2,2723925
$endgroup$
I am having trouble understanding the following statement:
3.20 Proposition, pg44: Let $D$ be a symmetric, odd graded elliptic operator on a graded vector bundle $S$ over a compact manifold. The element $[phi_D] in K(Bbb C) cong Bbb Z$ is the Fredholm index of the operator $D$.
What exactly are used in making sense of this statement:
Def 2.10, page 19. A Graded $*$-homomoprhism $phi_D:mathcalS rightarrow mathcalK(H)$ from spectral theorem.
Prop 3.17. There is an isomoprhism $$Phi:K(A) rightarrow [mathcalS, A otimes mathcalK(H)]$$
What the above two means is explain in my other post.
The proof goes as follows:
We have a homotopy of $*$-homomoprhisms $phi_s^-1D(f) = f(s^-1D)$. At $s=1$ we have $phi_D$ at $s=0$ we have the homomorphism of $f mapsto f(0)P$ where $P$ is projection onto the kernel of $D$.
How does one obtain continuity at $s=0$ - why is the resulting map as described?
This corresponds to the integer $dim(ker D cap H_+) - dim (ker D cap H_-)$.
How does one make this computation?
operator-algebras k-theory topological-k-theory
operator-algebras k-theory topological-k-theory
asked Mar 18 at 17:55
CL.CL.
2,2723925
asked Mar 18 at 17:55
CL.CL.
2,2723925
asked Mar 18 at 17:55
CL.CL.
2,2723925
asked Mar 18 at 17:55
CL.CL.
2,2723925
asked Mar 18 at 17:55
CL.CL.
2,2723925
2,2723925
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-frac12$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $lambda=+infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $mathbbR$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^-1D)$ are considered under the norm topology. Since
$$lim_sto 0 f(s^-1x)to delta_0(x)f(0)$$
As continuous functions over the spectrum of $D$. Also $delta_0(D)$ is the projection $P$ to the eigenspace for $lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $ker Dcap H_+$, $ker Dcap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have
$$ dim(ker Dcap H_+)=Tr(P_+),quad dim(ker Dcap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$
answered Apr 8 at 0:55
RuiRui
261
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-frac12$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $lambda=+infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $mathbbR$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^-1D)$ are considered under the norm topology. Since
$$lim_sto 0 f(s^-1x)to delta_0(x)f(0)$$
As continuous functions over the spectrum of $D$. Also $delta_0(D)$ is the projection $P$ to the eigenspace for $lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $ker Dcap H_+$, $ker Dcap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have
$$ dim(ker Dcap H_+)=Tr(P_+),quad dim(ker Dcap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$
answered Apr 8 at 0:55
RuiRui
261
$endgroup$
add a comment |
$begingroup$
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-frac12$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $lambda=+infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $mathbbR$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^-1D)$ are considered under the norm topology. Since
$$lim_sto 0 f(s^-1x)to delta_0(x)f(0)$$
As continuous functions over the spectrum of $D$. Also $delta_0(D)$ is the projection $P$ to the eigenspace for $lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $ker Dcap H_+$, $ker Dcap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have
$$ dim(ker Dcap H_+)=Tr(P_+),quad dim(ker Dcap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$
answered Apr 8 at 0:55
RuiRui
261
$endgroup$
add a comment |
$begingroup$
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-frac12$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $lambda=+infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $mathbbR$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^-1D)$ are considered under the norm topology. Since
$$lim_sto 0 f(s^-1x)to delta_0(x)f(0)$$
As continuous functions over the spectrum of $D$. Also $delta_0(D)$ is the projection $P$ to the eigenspace for $lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $ker Dcap H_+$, $ker Dcap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have
$$ dim(ker Dcap H_+)=Tr(P_+),quad dim(ker Dcap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$
answered Apr 8 at 0:55
RuiRui
261
$endgroup$
The operator $D$ here is special: If $D$ is elliptic on a compact manifold, it will have compact resolvant by Rellich's lemma: $(1+D^2)^-frac12$ is compact. Therefore the spectrum of $D$ are eigenvalues, they have no limit point except $lambda=+infty$. If $f(x)$ is a continuous function of spectrum of $D$, $f(D)$ is well-defined. This will contain something like $delta_0(x)$ who is 1 when $x=0$ and is 0 elsewhere. This is not continuous on $mathbbR$, but is continuous on the spectrum of $D$!
If $f$ vanishes at the infinity, $f(D)$ is bounded. The continuity of the operators $f(s^-1D)$ are considered under the norm topology. Since
$$lim_sto 0 f(s^-1x)to delta_0(x)f(0)$$
As continuous functions over the spectrum of $D$. Also $delta_0(D)$ is the projection $P$ to the eigenspace for $lambda=0$. We have the first limit of your question (1).
Also by the result of compact resolvant, the dimension of the eigenspaces are finite dimensional, Let $P_+ ,P_-$ be the projection on $ker Dcap H_+$, $ker Dcap H_-$ respectivly. They will be trace-class operators(as operators on $L^2(M)$. We have
$$ dim(ker Dcap H_+)=Tr(P_+),quad dim(ker Dcap H_-)=Tr(P_-)$$
We define a "supertrace" $Str(P):=Tr(P_+)-Tr(P_-)$.
We have $Index(D)=Str(P)$
answered Apr 8 at 0:55
RuiRui
261
answered Apr 8 at 0:55
RuiRui
261
answered Apr 8 at 0:55
RuiRui
261
answered Apr 8 at 0:55
RuiRui
261
261
add a comment |
add a comment |
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