Special elements in the $C^*$ algebra $A otimes mathcalK$.Explaining the group structure in homotopy classes of map $[X,K(H)]$$K$-Theory of operators I, Higson notes$K$ Theory of Operators II, Higson's notesShort exact sequence in $K_0$ of non unital rings.Containment of an element to an operator systemQuestion about special $C^*$-algebraCompact operators on $L^2(G)$ as a reduced cross product of $C_0(G)$ and $G$.$mathcalA+K$ is norm-closed where $mathcalA$ is a $C^*$-algebra and $K$ is the compact operators.Can $xy$ and $yx$ lie in different connected components of the group of invertible elements of an algebra?Two questions about positive elements in a C$^*$-algebraelements in $C^*$ algebraNormal u.c.p extension of Schur-multiplierInduced $K$-theory maps between $C^*$ algebras.Understanding the map from $K_0(A)$ to homotopy class of maps,

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Special elements in the $C^*$ algebra $A otimes mathcalK$.


Explaining the group structure in homotopy classes of map $[X,K(H)]$$K$-Theory of operators I, Higson notes$K$ Theory of Operators II, Higson's notesShort exact sequence in $K_0$ of non unital rings.Containment of an element to an operator systemQuestion about special $C^*$-algebraCompact operators on $L^2(G)$ as a reduced cross product of $C_0(G)$ and $G$.$mathcalA+K$ is norm-closed where $mathcalA$ is a $C^*$-algebra and $K$ is the compact operators.Can $xy$ and $yx$ lie in different connected components of the group of invertible elements of an algebra?Two questions about positive elements in a C$^*$-algebraelements in $C^*$ algebraNormal u.c.p extension of Schur-multiplierInduced $K$-theory maps between $C^*$ algebras.Understanding the map from $K_0(A)$ to homotopy class of maps,













4












$begingroup$



Context: Let $A$ be an ungraded (not necessarily unital) $C^*$ algebra. $mathcalK$ space of compact bounded operators on an infinite separable graded Hilbert space $H=H_0 oplus H_1$.
Consider the space
$$ A otimes mathcalK $$
Let us suppose there is a unique norm.





Edit: I replaced a large part of text which can be seen in history. For streamlining the post.




Claim 1' If we begin with a graded homomoprhism, $mathcalS rightarrow A otimes mathcalK$, then the unitary $u$ we obtain this way (via the Cayley transform) has the property that $alpha(u)=u^*$.



Claim 2: For any unital graded $C^*$ algebra $B$ containing $A otimes mathcalK$, consider the grading element,
$$
epsilon =
beginpmatrix
1 & 0 \
0 & -1
endpmatrix
$$

which grades $mathcalK$. Any skew unitary $u$ is equal to
$$p_epsilon = beginpmatrix
1 & 0 \
0 & 0
endpmatrix
$$

modulo $A otimes mathcalK$, i.e. $p_phi-p_epsilon in A otimes mathcalK$.





May someone elaborate the details? These are from page 43, proof of Prop 3.17 , Higson's notes.




Questions regarding Aweygan's reply



I am still confused:




  1. Why is it to say that $ bin B$ lies in $A otimes mathcalK$ precisely means that the scalar part of each $(b_ij)$ is zero. (This is more or less related to Q3) and pehraps Exercise 3.1 in page 37 of notes. I would like to understand a related question.


  2. $phi(f)$ is not a $2 times 2$ matrix here (?) i.e. why
    $$ phi(f) = (v_ij) in A otimes mathcalK?$$


  3. Elaboration on 2 and how Aweygan did this: how we define $tildephi$: first decompose a function $C(S^1)$ as $c+f$, where $f in C_0(Bbb R)$, $c$ is a constant. define
    $$ tildephi(c+f) = cI + phi(f) $$
    we embed $phi(f) in A otimes mathcalK rightarrow B$. But what is this embedding?











share|cite|improve this question











$endgroup$











  • $begingroup$
    For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
    $endgroup$
    – Aweygan
    Mar 7 at 17:31















4












$begingroup$



Context: Let $A$ be an ungraded (not necessarily unital) $C^*$ algebra. $mathcalK$ space of compact bounded operators on an infinite separable graded Hilbert space $H=H_0 oplus H_1$.
Consider the space
$$ A otimes mathcalK $$
Let us suppose there is a unique norm.





Edit: I replaced a large part of text which can be seen in history. For streamlining the post.




Claim 1' If we begin with a graded homomoprhism, $mathcalS rightarrow A otimes mathcalK$, then the unitary $u$ we obtain this way (via the Cayley transform) has the property that $alpha(u)=u^*$.



Claim 2: For any unital graded $C^*$ algebra $B$ containing $A otimes mathcalK$, consider the grading element,
$$
epsilon =
beginpmatrix
1 & 0 \
0 & -1
endpmatrix
$$

which grades $mathcalK$. Any skew unitary $u$ is equal to
$$p_epsilon = beginpmatrix
1 & 0 \
0 & 0
endpmatrix
$$

modulo $A otimes mathcalK$, i.e. $p_phi-p_epsilon in A otimes mathcalK$.





May someone elaborate the details? These are from page 43, proof of Prop 3.17 , Higson's notes.




Questions regarding Aweygan's reply



I am still confused:




  1. Why is it to say that $ bin B$ lies in $A otimes mathcalK$ precisely means that the scalar part of each $(b_ij)$ is zero. (This is more or less related to Q3) and pehraps Exercise 3.1 in page 37 of notes. I would like to understand a related question.


  2. $phi(f)$ is not a $2 times 2$ matrix here (?) i.e. why
    $$ phi(f) = (v_ij) in A otimes mathcalK?$$


  3. Elaboration on 2 and how Aweygan did this: how we define $tildephi$: first decompose a function $C(S^1)$ as $c+f$, where $f in C_0(Bbb R)$, $c$ is a constant. define
    $$ tildephi(c+f) = cI + phi(f) $$
    we embed $phi(f) in A otimes mathcalK rightarrow B$. But what is this embedding?











share|cite|improve this question











$endgroup$











  • $begingroup$
    For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
    $endgroup$
    – Aweygan
    Mar 7 at 17:31













4












4








4


1



$begingroup$



Context: Let $A$ be an ungraded (not necessarily unital) $C^*$ algebra. $mathcalK$ space of compact bounded operators on an infinite separable graded Hilbert space $H=H_0 oplus H_1$.
Consider the space
$$ A otimes mathcalK $$
Let us suppose there is a unique norm.





Edit: I replaced a large part of text which can be seen in history. For streamlining the post.




Claim 1' If we begin with a graded homomoprhism, $mathcalS rightarrow A otimes mathcalK$, then the unitary $u$ we obtain this way (via the Cayley transform) has the property that $alpha(u)=u^*$.



Claim 2: For any unital graded $C^*$ algebra $B$ containing $A otimes mathcalK$, consider the grading element,
$$
epsilon =
beginpmatrix
1 & 0 \
0 & -1
endpmatrix
$$

which grades $mathcalK$. Any skew unitary $u$ is equal to
$$p_epsilon = beginpmatrix
1 & 0 \
0 & 0
endpmatrix
$$

modulo $A otimes mathcalK$, i.e. $p_phi-p_epsilon in A otimes mathcalK$.





May someone elaborate the details? These are from page 43, proof of Prop 3.17 , Higson's notes.




Questions regarding Aweygan's reply



I am still confused:




  1. Why is it to say that $ bin B$ lies in $A otimes mathcalK$ precisely means that the scalar part of each $(b_ij)$ is zero. (This is more or less related to Q3) and pehraps Exercise 3.1 in page 37 of notes. I would like to understand a related question.


  2. $phi(f)$ is not a $2 times 2$ matrix here (?) i.e. why
    $$ phi(f) = (v_ij) in A otimes mathcalK?$$


  3. Elaboration on 2 and how Aweygan did this: how we define $tildephi$: first decompose a function $C(S^1)$ as $c+f$, where $f in C_0(Bbb R)$, $c$ is a constant. define
    $$ tildephi(c+f) = cI + phi(f) $$
    we embed $phi(f) in A otimes mathcalK rightarrow B$. But what is this embedding?











share|cite|improve this question











$endgroup$





Context: Let $A$ be an ungraded (not necessarily unital) $C^*$ algebra. $mathcalK$ space of compact bounded operators on an infinite separable graded Hilbert space $H=H_0 oplus H_1$.
Consider the space
$$ A otimes mathcalK $$
Let us suppose there is a unique norm.





Edit: I replaced a large part of text which can be seen in history. For streamlining the post.




Claim 1' If we begin with a graded homomoprhism, $mathcalS rightarrow A otimes mathcalK$, then the unitary $u$ we obtain this way (via the Cayley transform) has the property that $alpha(u)=u^*$.



Claim 2: For any unital graded $C^*$ algebra $B$ containing $A otimes mathcalK$, consider the grading element,
$$
epsilon =
beginpmatrix
1 & 0 \
0 & -1
endpmatrix
$$

which grades $mathcalK$. Any skew unitary $u$ is equal to
$$p_epsilon = beginpmatrix
1 & 0 \
0 & 0
endpmatrix
$$

modulo $A otimes mathcalK$, i.e. $p_phi-p_epsilon in A otimes mathcalK$.





May someone elaborate the details? These are from page 43, proof of Prop 3.17 , Higson's notes.




Questions regarding Aweygan's reply



I am still confused:




  1. Why is it to say that $ bin B$ lies in $A otimes mathcalK$ precisely means that the scalar part of each $(b_ij)$ is zero. (This is more or less related to Q3) and pehraps Exercise 3.1 in page 37 of notes. I would like to understand a related question.


  2. $phi(f)$ is not a $2 times 2$ matrix here (?) i.e. why
    $$ phi(f) = (v_ij) in A otimes mathcalK?$$


  3. Elaboration on 2 and how Aweygan did this: how we define $tildephi$: first decompose a function $C(S^1)$ as $c+f$, where $f in C_0(Bbb R)$, $c$ is a constant. define
    $$ tildephi(c+f) = cI + phi(f) $$
    we embed $phi(f) in A otimes mathcalK rightarrow B$. But what is this embedding?








abstract-algebra functional-analysis operator-theory operator-algebras k-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 28 at 16:27







CL.

















asked Mar 7 at 11:36









CL.CL.

2,2672925




2,2672925











  • $begingroup$
    For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
    $endgroup$
    – Aweygan
    Mar 7 at 17:31
















  • $begingroup$
    For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
    $endgroup$
    – Aweygan
    Mar 7 at 17:31















$begingroup$
For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
$endgroup$
– Aweygan
Mar 7 at 17:31




$begingroup$
For claim 1, it looks like you might be misenterpreting the data. I suggest rereading this material. Otherwise, I'll post an answer once I've thoroughly understood claim 2.
$endgroup$
– Aweygan
Mar 7 at 17:31










1 Answer
1






active

oldest

votes


















2












$begingroup$

For claim 1, the information copied here is not quite what they have stated in the book.




Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $alpha:Ato A$. A unitary $uin A$ is called a skew-unitary if $alpha(u)=u^*$.

If the grading is internal, i.e., there is some self-adjoint unitary $varepsilonin A$ such that $alpha(x)=varepsilon xvarepsilon$ for all $xin A$, then the map from skew unitaries to projections given by $umapsto frac12(1+uvarepsilon)$ is a bijection.




With this information, it should be clear why $frac12(1+uvarepsilon)$ is a projection if $varepsilon$ is a self-adjoint unitary and $varepsilon uvarepsilon=u^*$.




For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_epsilon$, only a very special one. In this section, $phi:mathcal Sto Aotimesmathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $phi$ induces a unital $*$-homomorphism $tildephi$ from $C(S^1)$ to the unitization of $Aotimesmathcal K$. The unitary in question is then $tildephi(z)$, where $z:S^1tomathbb C$ is the inclusion map.



Using the grading on $mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2times 2$-matrices with entries in $widetildeAotimesmathcal K$ (the unitization of $Aotimesmathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_ij)in B$ lies in $Aotimesmathcal K$ precisely means that the scalar part of each entry $b_ij$ is zero.



From the graded $*$-homomorphism $phi:C_0(mathbb R)to Aotimesmathcal K$, we obtain a unital $*$-homomorphism $tildephi:widetildeC_0(mathbb R)to B$ (by mapping units to units, and everything else by $phi$). Note that $widetildeC_0(mathbb R)=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $fin C_0(mathbb R)$, and
$$tildephi(u)=beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrix$$
where $phi(f)=(v_ij)in Aotimesmathcal K$. Then we have
$$p_phi=frac12(1+tildephi(u)epsilon=frac12left(beginpmatrix1&0\0&1endpmatrix+beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrixbeginpmatrix1&0\0&-1endpmatrixright)=beginpmatrix1+fracv_112&frac-v_122\fracv_212&frac-v_222endpmatrix,$$
so that
$$p_phi-p_epsilon=beginpmatrixfracv_112&frac-v_122\fracv_212&frac-v_222endpmatrixin Aotimesmathcal K.$$




Further Questions:



  1. Note that $Aotimesmathcal K$ is isomorphic to $M_2(Aotimesmathcal K)$, by decomposing the Hilbert space $H$ that $mathcal K$ acts on into a direct sum $H=H_0oplus H_1$ (This is also how the grading on $mathcal K$ is defined). So when I say an element $(b_ij)$ of $B$ lies in $Aotimesmathcal K$ when the scalar part of each $b_ij$ is zero, I really mean that $(b_ij)$ lies in $M_2(Aotimesmathcal K)$.


  2. As I said above, $Aotimesmathcal K$ is graded so that it looks like $M_2(Aotimesmathcal K)$. Thus the homomorphism $phi:mathcal Sto Aotimesmathcal K$ looks like a homomorphism $mathcal Sto M_2(Aotimesmathcal K)$.


  3. How I have $B$ defined, a typical element of $B$ looks like a $2times 2$ matrix $(b_ij)=(a_ij+lambda_ij)$, where $a_ijin Aotimesmathcal K$ and $lambda_ijinmathbb C$. The embedding $Aotimesmathcal Kto B$ is just $(a_ij)mapsto(a_ij+0)$, with scalar part $0$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
    $endgroup$
    – Aweygan
    Mar 17 at 14:33










  • $begingroup$
    I think I have it. I'll write up what I have; let me know if you have any questions.
    $endgroup$
    – Aweygan
    Mar 19 at 17:42










  • $begingroup$
    The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
    $endgroup$
    – Aweygan
    Mar 21 at 17:03










  • $begingroup$
    Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
    $endgroup$
    – CL.
    Mar 25 at 21:49











  • $begingroup$
    I edited my answer a few days ago. Was this not what you were looking for?
    $endgroup$
    – Aweygan
    Mar 25 at 22:35











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For claim 1, the information copied here is not quite what they have stated in the book.




Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $alpha:Ato A$. A unitary $uin A$ is called a skew-unitary if $alpha(u)=u^*$.

If the grading is internal, i.e., there is some self-adjoint unitary $varepsilonin A$ such that $alpha(x)=varepsilon xvarepsilon$ for all $xin A$, then the map from skew unitaries to projections given by $umapsto frac12(1+uvarepsilon)$ is a bijection.




With this information, it should be clear why $frac12(1+uvarepsilon)$ is a projection if $varepsilon$ is a self-adjoint unitary and $varepsilon uvarepsilon=u^*$.




For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_epsilon$, only a very special one. In this section, $phi:mathcal Sto Aotimesmathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $phi$ induces a unital $*$-homomorphism $tildephi$ from $C(S^1)$ to the unitization of $Aotimesmathcal K$. The unitary in question is then $tildephi(z)$, where $z:S^1tomathbb C$ is the inclusion map.



Using the grading on $mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2times 2$-matrices with entries in $widetildeAotimesmathcal K$ (the unitization of $Aotimesmathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_ij)in B$ lies in $Aotimesmathcal K$ precisely means that the scalar part of each entry $b_ij$ is zero.



From the graded $*$-homomorphism $phi:C_0(mathbb R)to Aotimesmathcal K$, we obtain a unital $*$-homomorphism $tildephi:widetildeC_0(mathbb R)to B$ (by mapping units to units, and everything else by $phi$). Note that $widetildeC_0(mathbb R)=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $fin C_0(mathbb R)$, and
$$tildephi(u)=beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrix$$
where $phi(f)=(v_ij)in Aotimesmathcal K$. Then we have
$$p_phi=frac12(1+tildephi(u)epsilon=frac12left(beginpmatrix1&0\0&1endpmatrix+beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrixbeginpmatrix1&0\0&-1endpmatrixright)=beginpmatrix1+fracv_112&frac-v_122\fracv_212&frac-v_222endpmatrix,$$
so that
$$p_phi-p_epsilon=beginpmatrixfracv_112&frac-v_122\fracv_212&frac-v_222endpmatrixin Aotimesmathcal K.$$




Further Questions:



  1. Note that $Aotimesmathcal K$ is isomorphic to $M_2(Aotimesmathcal K)$, by decomposing the Hilbert space $H$ that $mathcal K$ acts on into a direct sum $H=H_0oplus H_1$ (This is also how the grading on $mathcal K$ is defined). So when I say an element $(b_ij)$ of $B$ lies in $Aotimesmathcal K$ when the scalar part of each $b_ij$ is zero, I really mean that $(b_ij)$ lies in $M_2(Aotimesmathcal K)$.


  2. As I said above, $Aotimesmathcal K$ is graded so that it looks like $M_2(Aotimesmathcal K)$. Thus the homomorphism $phi:mathcal Sto Aotimesmathcal K$ looks like a homomorphism $mathcal Sto M_2(Aotimesmathcal K)$.


  3. How I have $B$ defined, a typical element of $B$ looks like a $2times 2$ matrix $(b_ij)=(a_ij+lambda_ij)$, where $a_ijin Aotimesmathcal K$ and $lambda_ijinmathbb C$. The embedding $Aotimesmathcal Kto B$ is just $(a_ij)mapsto(a_ij+0)$, with scalar part $0$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
    $endgroup$
    – Aweygan
    Mar 17 at 14:33










  • $begingroup$
    I think I have it. I'll write up what I have; let me know if you have any questions.
    $endgroup$
    – Aweygan
    Mar 19 at 17:42










  • $begingroup$
    The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
    $endgroup$
    – Aweygan
    Mar 21 at 17:03










  • $begingroup$
    Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
    $endgroup$
    – CL.
    Mar 25 at 21:49











  • $begingroup$
    I edited my answer a few days ago. Was this not what you were looking for?
    $endgroup$
    – Aweygan
    Mar 25 at 22:35















2












$begingroup$

For claim 1, the information copied here is not quite what they have stated in the book.




Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $alpha:Ato A$. A unitary $uin A$ is called a skew-unitary if $alpha(u)=u^*$.

If the grading is internal, i.e., there is some self-adjoint unitary $varepsilonin A$ such that $alpha(x)=varepsilon xvarepsilon$ for all $xin A$, then the map from skew unitaries to projections given by $umapsto frac12(1+uvarepsilon)$ is a bijection.




With this information, it should be clear why $frac12(1+uvarepsilon)$ is a projection if $varepsilon$ is a self-adjoint unitary and $varepsilon uvarepsilon=u^*$.




For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_epsilon$, only a very special one. In this section, $phi:mathcal Sto Aotimesmathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $phi$ induces a unital $*$-homomorphism $tildephi$ from $C(S^1)$ to the unitization of $Aotimesmathcal K$. The unitary in question is then $tildephi(z)$, where $z:S^1tomathbb C$ is the inclusion map.



Using the grading on $mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2times 2$-matrices with entries in $widetildeAotimesmathcal K$ (the unitization of $Aotimesmathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_ij)in B$ lies in $Aotimesmathcal K$ precisely means that the scalar part of each entry $b_ij$ is zero.



From the graded $*$-homomorphism $phi:C_0(mathbb R)to Aotimesmathcal K$, we obtain a unital $*$-homomorphism $tildephi:widetildeC_0(mathbb R)to B$ (by mapping units to units, and everything else by $phi$). Note that $widetildeC_0(mathbb R)=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $fin C_0(mathbb R)$, and
$$tildephi(u)=beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrix$$
where $phi(f)=(v_ij)in Aotimesmathcal K$. Then we have
$$p_phi=frac12(1+tildephi(u)epsilon=frac12left(beginpmatrix1&0\0&1endpmatrix+beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrixbeginpmatrix1&0\0&-1endpmatrixright)=beginpmatrix1+fracv_112&frac-v_122\fracv_212&frac-v_222endpmatrix,$$
so that
$$p_phi-p_epsilon=beginpmatrixfracv_112&frac-v_122\fracv_212&frac-v_222endpmatrixin Aotimesmathcal K.$$




Further Questions:



  1. Note that $Aotimesmathcal K$ is isomorphic to $M_2(Aotimesmathcal K)$, by decomposing the Hilbert space $H$ that $mathcal K$ acts on into a direct sum $H=H_0oplus H_1$ (This is also how the grading on $mathcal K$ is defined). So when I say an element $(b_ij)$ of $B$ lies in $Aotimesmathcal K$ when the scalar part of each $b_ij$ is zero, I really mean that $(b_ij)$ lies in $M_2(Aotimesmathcal K)$.


  2. As I said above, $Aotimesmathcal K$ is graded so that it looks like $M_2(Aotimesmathcal K)$. Thus the homomorphism $phi:mathcal Sto Aotimesmathcal K$ looks like a homomorphism $mathcal Sto M_2(Aotimesmathcal K)$.


  3. How I have $B$ defined, a typical element of $B$ looks like a $2times 2$ matrix $(b_ij)=(a_ij+lambda_ij)$, where $a_ijin Aotimesmathcal K$ and $lambda_ijinmathbb C$. The embedding $Aotimesmathcal Kto B$ is just $(a_ij)mapsto(a_ij+0)$, with scalar part $0$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
    $endgroup$
    – Aweygan
    Mar 17 at 14:33










  • $begingroup$
    I think I have it. I'll write up what I have; let me know if you have any questions.
    $endgroup$
    – Aweygan
    Mar 19 at 17:42










  • $begingroup$
    The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
    $endgroup$
    – Aweygan
    Mar 21 at 17:03










  • $begingroup$
    Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
    $endgroup$
    – CL.
    Mar 25 at 21:49











  • $begingroup$
    I edited my answer a few days ago. Was this not what you were looking for?
    $endgroup$
    – Aweygan
    Mar 25 at 22:35













2












2








2





$begingroup$

For claim 1, the information copied here is not quite what they have stated in the book.




Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $alpha:Ato A$. A unitary $uin A$ is called a skew-unitary if $alpha(u)=u^*$.

If the grading is internal, i.e., there is some self-adjoint unitary $varepsilonin A$ such that $alpha(x)=varepsilon xvarepsilon$ for all $xin A$, then the map from skew unitaries to projections given by $umapsto frac12(1+uvarepsilon)$ is a bijection.




With this information, it should be clear why $frac12(1+uvarepsilon)$ is a projection if $varepsilon$ is a self-adjoint unitary and $varepsilon uvarepsilon=u^*$.




For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_epsilon$, only a very special one. In this section, $phi:mathcal Sto Aotimesmathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $phi$ induces a unital $*$-homomorphism $tildephi$ from $C(S^1)$ to the unitization of $Aotimesmathcal K$. The unitary in question is then $tildephi(z)$, where $z:S^1tomathbb C$ is the inclusion map.



Using the grading on $mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2times 2$-matrices with entries in $widetildeAotimesmathcal K$ (the unitization of $Aotimesmathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_ij)in B$ lies in $Aotimesmathcal K$ precisely means that the scalar part of each entry $b_ij$ is zero.



From the graded $*$-homomorphism $phi:C_0(mathbb R)to Aotimesmathcal K$, we obtain a unital $*$-homomorphism $tildephi:widetildeC_0(mathbb R)to B$ (by mapping units to units, and everything else by $phi$). Note that $widetildeC_0(mathbb R)=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $fin C_0(mathbb R)$, and
$$tildephi(u)=beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrix$$
where $phi(f)=(v_ij)in Aotimesmathcal K$. Then we have
$$p_phi=frac12(1+tildephi(u)epsilon=frac12left(beginpmatrix1&0\0&1endpmatrix+beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrixbeginpmatrix1&0\0&-1endpmatrixright)=beginpmatrix1+fracv_112&frac-v_122\fracv_212&frac-v_222endpmatrix,$$
so that
$$p_phi-p_epsilon=beginpmatrixfracv_112&frac-v_122\fracv_212&frac-v_222endpmatrixin Aotimesmathcal K.$$




Further Questions:



  1. Note that $Aotimesmathcal K$ is isomorphic to $M_2(Aotimesmathcal K)$, by decomposing the Hilbert space $H$ that $mathcal K$ acts on into a direct sum $H=H_0oplus H_1$ (This is also how the grading on $mathcal K$ is defined). So when I say an element $(b_ij)$ of $B$ lies in $Aotimesmathcal K$ when the scalar part of each $b_ij$ is zero, I really mean that $(b_ij)$ lies in $M_2(Aotimesmathcal K)$.


  2. As I said above, $Aotimesmathcal K$ is graded so that it looks like $M_2(Aotimesmathcal K)$. Thus the homomorphism $phi:mathcal Sto Aotimesmathcal K$ looks like a homomorphism $mathcal Sto M_2(Aotimesmathcal K)$.


  3. How I have $B$ defined, a typical element of $B$ looks like a $2times 2$ matrix $(b_ij)=(a_ij+lambda_ij)$, where $a_ijin Aotimesmathcal K$ and $lambda_ijinmathbb C$. The embedding $Aotimesmathcal Kto B$ is just $(a_ij)mapsto(a_ij+0)$, with scalar part $0$.






share|cite|improve this answer











$endgroup$



For claim 1, the information copied here is not quite what they have stated in the book.




Suppose $A$ is a graded unital $C^*$-algebra, with the grading given by a $*$-automorphism $alpha:Ato A$. A unitary $uin A$ is called a skew-unitary if $alpha(u)=u^*$.

If the grading is internal, i.e., there is some self-adjoint unitary $varepsilonin A$ such that $alpha(x)=varepsilon xvarepsilon$ for all $xin A$, then the map from skew unitaries to projections given by $umapsto frac12(1+uvarepsilon)$ is a bijection.




With this information, it should be clear why $frac12(1+uvarepsilon)$ is a projection if $varepsilon$ is a self-adjoint unitary and $varepsilon uvarepsilon=u^*$.




For claim 2, the authors aren't claiming that any skew unitary is equivalent $p_epsilon$, only a very special one. In this section, $phi:mathcal Sto Aotimesmathcal K$ is a graded $*$-homomorphism. Via the Cayley transform, $phi$ induces a unital $*$-homomorphism $tildephi$ from $C(S^1)$ to the unitization of $Aotimesmathcal K$. The unitary in question is then $tildephi(z)$, where $z:S^1tomathbb C$ is the inclusion map.



Using the grading on $mathcal K$, we can consider the algebra $B$ in question as the algebra of all $2times 2$-matrices with entries in $widetildeAotimesmathcal K$ (the unitization of $Aotimesmathcal K$), graded by diagonal matrices (even part) and off-diagonal matrices (odd part). Then to say that $b=(b_ij)in B$ lies in $Aotimesmathcal K$ precisely means that the scalar part of each entry $b_ij$ is zero.



From the graded $*$-homomorphism $phi:C_0(mathbb R)to Aotimesmathcal K$, we obtain a unital $*$-homomorphism $tildephi:widetildeC_0(mathbb R)to B$ (by mapping units to units, and everything else by $phi$). Note that $widetildeC_0(mathbb R)=C(S^1)$ is generated by a single unitary $u$. Then $u=1+f$ for some $fin C_0(mathbb R)$, and
$$tildephi(u)=beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrix$$
where $phi(f)=(v_ij)in Aotimesmathcal K$. Then we have
$$p_phi=frac12(1+tildephi(u)epsilon=frac12left(beginpmatrix1&0\0&1endpmatrix+beginpmatrix1+v_11&v_12\v_21&1+v_22 endpmatrixbeginpmatrix1&0\0&-1endpmatrixright)=beginpmatrix1+fracv_112&frac-v_122\fracv_212&frac-v_222endpmatrix,$$
so that
$$p_phi-p_epsilon=beginpmatrixfracv_112&frac-v_122\fracv_212&frac-v_222endpmatrixin Aotimesmathcal K.$$




Further Questions:



  1. Note that $Aotimesmathcal K$ is isomorphic to $M_2(Aotimesmathcal K)$, by decomposing the Hilbert space $H$ that $mathcal K$ acts on into a direct sum $H=H_0oplus H_1$ (This is also how the grading on $mathcal K$ is defined). So when I say an element $(b_ij)$ of $B$ lies in $Aotimesmathcal K$ when the scalar part of each $b_ij$ is zero, I really mean that $(b_ij)$ lies in $M_2(Aotimesmathcal K)$.


  2. As I said above, $Aotimesmathcal K$ is graded so that it looks like $M_2(Aotimesmathcal K)$. Thus the homomorphism $phi:mathcal Sto Aotimesmathcal K$ looks like a homomorphism $mathcal Sto M_2(Aotimesmathcal K)$.


  3. How I have $B$ defined, a typical element of $B$ looks like a $2times 2$ matrix $(b_ij)=(a_ij+lambda_ij)$, where $a_ijin Aotimesmathcal K$ and $lambda_ijinmathbb C$. The embedding $Aotimesmathcal Kto B$ is just $(a_ij)mapsto(a_ij+0)$, with scalar part $0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 24 at 15:20

























answered Mar 11 at 18:27









AweyganAweygan

14.7k21442




14.7k21442











  • $begingroup$
    The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
    $endgroup$
    – Aweygan
    Mar 17 at 14:33










  • $begingroup$
    I think I have it. I'll write up what I have; let me know if you have any questions.
    $endgroup$
    – Aweygan
    Mar 19 at 17:42










  • $begingroup$
    The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
    $endgroup$
    – Aweygan
    Mar 21 at 17:03










  • $begingroup$
    Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
    $endgroup$
    – CL.
    Mar 25 at 21:49











  • $begingroup$
    I edited my answer a few days ago. Was this not what you were looking for?
    $endgroup$
    – Aweygan
    Mar 25 at 22:35
















  • $begingroup$
    The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
    $endgroup$
    – Aweygan
    Mar 17 at 14:33










  • $begingroup$
    I think I have it. I'll write up what I have; let me know if you have any questions.
    $endgroup$
    – Aweygan
    Mar 19 at 17:42










  • $begingroup$
    The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
    $endgroup$
    – Aweygan
    Mar 21 at 17:03










  • $begingroup$
    Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
    $endgroup$
    – CL.
    Mar 25 at 21:49











  • $begingroup$
    I edited my answer a few days ago. Was this not what you were looking for?
    $endgroup$
    – Aweygan
    Mar 25 at 22:35















$begingroup$
The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
$endgroup$
– Aweygan
Mar 17 at 14:33




$begingroup$
The grading of $mathcal S$ is by even and odd functions. You can consider $B$ to be the unitization of $M_2(Aotimesmathcal K)$, graded by diagonal matrices (even part) and off-diagonal matrices (odd part). They don't claim that $u-p_varepsilonin Aotimesmathcal K$, but $p_phi-p_varepsilon$ is, where $p_varepsilon=frac12(1+uvarepsilon)$. I'm still trying to understand this myself, and I will edit my answer once I see this.
$endgroup$
– Aweygan
Mar 17 at 14:33












$begingroup$
I think I have it. I'll write up what I have; let me know if you have any questions.
$endgroup$
– Aweygan
Mar 19 at 17:42




$begingroup$
I think I have it. I'll write up what I have; let me know if you have any questions.
$endgroup$
– Aweygan
Mar 19 at 17:42












$begingroup$
The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
$endgroup$
– Aweygan
Mar 21 at 17:03




$begingroup$
The unitary in question is the function $z:S^1tomathbb C$ given by inclusion: $z(e^itheta)=e^itheta$. When we say that $C(S^1)$ is generated by $z$, we mean that the $*$-subalgebra of all polynomials in $z$ and $z^*$ is dense in $C(S^1)$ (this follows from the Stone-Weierstrass theorem).
$endgroup$
– Aweygan
Mar 21 at 17:03












$begingroup$
Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
$endgroup$
– CL.
Mar 25 at 21:49





$begingroup$
Hi Aweygan, I fully understand you may be busy, but I would be grateful, if you could comment briefly on my new concerns - thank you so much!
$endgroup$
– CL.
Mar 25 at 21:49













$begingroup$
I edited my answer a few days ago. Was this not what you were looking for?
$endgroup$
– Aweygan
Mar 25 at 22:35




$begingroup$
I edited my answer a few days ago. Was this not what you were looking for?
$endgroup$
– Aweygan
Mar 25 at 22:35

















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