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How to use SNAG theorem to find the unitary representations of the additive group $mathbb{R}^n$?
$begingroup$
Consider the additive group $mathbb{R}^n$. It is an abelian, locally compact group. The SNAG theorem applies. That theorem says:
SNAG Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $mathscr{H}$. Then there exists on the character group $hat{G}$ a spectral measure $E$ such that $$T(x)=int_{hat{G}}langle hat{x},xrangle dE(hat{x})$$
It is said that for $mathbb{R}^n$ there are $n$ strongly commuting self-adjoint operators $Y_1,dots, Y_n$ with
$$T_x=prod_{k=1}^nexp(i x_k Y_k)$$
Now, how can this be derived?
I think the starting point is to find the character group. Since a character is a continuous group homomorphism $chi : Gto S^1$ it is $chi(g)=e^{itheta(g)}$ for $theta : Gto mathbb{R}$ a continuous function with $theta(gh)=theta(g)+theta(h)$.
For the $G = mathbb{R}^n$ the function $theta$ is continuous and satisfies $theta(g+h)=theta(g)+theta(h)$ so it is linear.
Hence fixing a basis ${e_i}$ of $mathbb{R}^n$ with dual basis ${omega^i}$ expanding $theta = sum theta_i omega^i$ and $g =sum g^ie_i$ we have
$$T_g=int_{(mathbb{R}^n)^ast}exp(isum_ktheta_k g^k) dE(theta)=int_{(mathbb{R}^n)^ast}prod exp(i theta_k g^k)dE(theta).$$
I think now we would need to split the integral as a product of integrals and then identify $Y_i$. But how this is done? Why the integral can be split as a product? How the $Y_i$ are identified?
How, rigorously, the SNAG theorem gives the unitary representations of $mathbb{R}^n$ in that form?
group-theory functional-analysis measure-theory representation-theory hilbert-spaces
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
$endgroup$
add a comment |
$begingroup$
Consider the additive group $mathbb{R}^n$. It is an abelian, locally compact group. The SNAG theorem applies. That theorem says:
SNAG Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $mathscr{H}$. Then there exists on the character group $hat{G}$ a spectral measure $E$ such that $$T(x)=int_{hat{G}}langle hat{x},xrangle dE(hat{x})$$
It is said that for $mathbb{R}^n$ there are $n$ strongly commuting self-adjoint operators $Y_1,dots, Y_n$ with
$$T_x=prod_{k=1}^nexp(i x_k Y_k)$$
Now, how can this be derived?
I think the starting point is to find the character group. Since a character is a continuous group homomorphism $chi : Gto S^1$ it is $chi(g)=e^{itheta(g)}$ for $theta : Gto mathbb{R}$ a continuous function with $theta(gh)=theta(g)+theta(h)$.
For the $G = mathbb{R}^n$ the function $theta$ is continuous and satisfies $theta(g+h)=theta(g)+theta(h)$ so it is linear.
Hence fixing a basis ${e_i}$ of $mathbb{R}^n$ with dual basis ${omega^i}$ expanding $theta = sum theta_i omega^i$ and $g =sum g^ie_i$ we have
$$T_g=int_{(mathbb{R}^n)^ast}exp(isum_ktheta_k g^k) dE(theta)=int_{(mathbb{R}^n)^ast}prod exp(i theta_k g^k)dE(theta).$$
I think now we would need to split the integral as a product of integrals and then identify $Y_i$. But how this is done? Why the integral can be split as a product? How the $Y_i$ are identified?
How, rigorously, the SNAG theorem gives the unitary representations of $mathbb{R}^n$ in that form?
group-theory functional-analysis measure-theory representation-theory hilbert-spaces
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
$endgroup$
add a comment |
$begingroup$
Consider the additive group $mathbb{R}^n$. It is an abelian, locally compact group. The SNAG theorem applies. That theorem says:
SNAG Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $mathscr{H}$. Then there exists on the character group $hat{G}$ a spectral measure $E$ such that $$T(x)=int_{hat{G}}langle hat{x},xrangle dE(hat{x})$$
It is said that for $mathbb{R}^n$ there are $n$ strongly commuting self-adjoint operators $Y_1,dots, Y_n$ with
$$T_x=prod_{k=1}^nexp(i x_k Y_k)$$
Now, how can this be derived?
I think the starting point is to find the character group. Since a character is a continuous group homomorphism $chi : Gto S^1$ it is $chi(g)=e^{itheta(g)}$ for $theta : Gto mathbb{R}$ a continuous function with $theta(gh)=theta(g)+theta(h)$.
For the $G = mathbb{R}^n$ the function $theta$ is continuous and satisfies $theta(g+h)=theta(g)+theta(h)$ so it is linear.
Hence fixing a basis ${e_i}$ of $mathbb{R}^n$ with dual basis ${omega^i}$ expanding $theta = sum theta_i omega^i$ and $g =sum g^ie_i$ we have
$$T_g=int_{(mathbb{R}^n)^ast}exp(isum_ktheta_k g^k) dE(theta)=int_{(mathbb{R}^n)^ast}prod exp(i theta_k g^k)dE(theta).$$
I think now we would need to split the integral as a product of integrals and then identify $Y_i$. But how this is done? Why the integral can be split as a product? How the $Y_i$ are identified?
How, rigorously, the SNAG theorem gives the unitary representations of $mathbb{R}^n$ in that form?
group-theory functional-analysis measure-theory representation-theory hilbert-spaces
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
$endgroup$
Consider the additive group $mathbb{R}^n$. It is an abelian, locally compact group. The SNAG theorem applies. That theorem says:
SNAG Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $mathscr{H}$. Then there exists on the character group $hat{G}$ a spectral measure $E$ such that $$T(x)=int_{hat{G}}langle hat{x},xrangle dE(hat{x})$$
It is said that for $mathbb{R}^n$ there are $n$ strongly commuting self-adjoint operators $Y_1,dots, Y_n$ with
$$T_x=prod_{k=1}^nexp(i x_k Y_k)$$
Now, how can this be derived?
I think the starting point is to find the character group. Since a character is a continuous group homomorphism $chi : Gto S^1$ it is $chi(g)=e^{itheta(g)}$ for $theta : Gto mathbb{R}$ a continuous function with $theta(gh)=theta(g)+theta(h)$.
For the $G = mathbb{R}^n$ the function $theta$ is continuous and satisfies $theta(g+h)=theta(g)+theta(h)$ so it is linear.
Hence fixing a basis ${e_i}$ of $mathbb{R}^n$ with dual basis ${omega^i}$ expanding $theta = sum theta_i omega^i$ and $g =sum g^ie_i$ we have
$$T_g=int_{(mathbb{R}^n)^ast}exp(isum_ktheta_k g^k) dE(theta)=int_{(mathbb{R}^n)^ast}prod exp(i theta_k g^k)dE(theta).$$
I think now we would need to split the integral as a product of integrals and then identify $Y_i$. But how this is done? Why the integral can be split as a product? How the $Y_i$ are identified?
How, rigorously, the SNAG theorem gives the unitary representations of $mathbb{R}^n$ in that form?
group-theory functional-analysis measure-theory representation-theory hilbert-spaces
group-theory functional-analysis measure-theory representation-theory hilbert-spaces
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
edited Jan 28 at 16:03
user1620696
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
asked Jan 28 at 14:33
user1620696user1620696
11.7k742118
11.7k742118
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Denote by $pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $mathbb{R}^n$, then the operators $Y_i$ defined by
begin{align*}
D(Y_k)&=left{xiinmathscr{H} : int pi_k(theta)^2,dlangle E(theta)xi,xirangle<inftyright},\
Y_kxi&=int pi_k(theta),dE(theta) xi
end{align*}
are self-adjoint with spectral measure $E_k=(pi_k)_#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $fmapsto int f,dE$ is a $ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $mathscr{H}$: For simple functions $f=sum_j a_j 1_{A_j}$ and $g=sum_k b_k 1_{B_k}$ one has
$$
int fg,dE=int sum_{j,k}a_j b_k 1_{A_jcap B_k},dE=sum_{j,k}a_j b_k E(A_j)E( B_k)=left(sum_j a_j E(A_j)right)left(sum_k b_k E(B_k)right)=left(int f,dEright)left(int g,dEright).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
int prod_{k=1}^n f_k(theta_k),dE(theta)=prod_{k=1}^n int f_k(theta_k),dE(theta)=prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,dots,f_n$.
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
$endgroup$
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
add a comment |
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1 Answer
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$begingroup$
Denote by $pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $mathbb{R}^n$, then the operators $Y_i$ defined by
begin{align*}
D(Y_k)&=left{xiinmathscr{H} : int pi_k(theta)^2,dlangle E(theta)xi,xirangle<inftyright},\
Y_kxi&=int pi_k(theta),dE(theta) xi
end{align*}
are self-adjoint with spectral measure $E_k=(pi_k)_#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $fmapsto int f,dE$ is a $ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $mathscr{H}$: For simple functions $f=sum_j a_j 1_{A_j}$ and $g=sum_k b_k 1_{B_k}$ one has
$$
int fg,dE=int sum_{j,k}a_j b_k 1_{A_jcap B_k},dE=sum_{j,k}a_j b_k E(A_j)E( B_k)=left(sum_j a_j E(A_j)right)left(sum_k b_k E(B_k)right)=left(int f,dEright)left(int g,dEright).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
int prod_{k=1}^n f_k(theta_k),dE(theta)=prod_{k=1}^n int f_k(theta_k),dE(theta)=prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,dots,f_n$.
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
$endgroup$
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
add a comment |
$begingroup$
Denote by $pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $mathbb{R}^n$, then the operators $Y_i$ defined by
begin{align*}
D(Y_k)&=left{xiinmathscr{H} : int pi_k(theta)^2,dlangle E(theta)xi,xirangle<inftyright},\
Y_kxi&=int pi_k(theta),dE(theta) xi
end{align*}
are self-adjoint with spectral measure $E_k=(pi_k)_#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $fmapsto int f,dE$ is a $ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $mathscr{H}$: For simple functions $f=sum_j a_j 1_{A_j}$ and $g=sum_k b_k 1_{B_k}$ one has
$$
int fg,dE=int sum_{j,k}a_j b_k 1_{A_jcap B_k},dE=sum_{j,k}a_j b_k E(A_j)E( B_k)=left(sum_j a_j E(A_j)right)left(sum_k b_k E(B_k)right)=left(int f,dEright)left(int g,dEright).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
int prod_{k=1}^n f_k(theta_k),dE(theta)=prod_{k=1}^n int f_k(theta_k),dE(theta)=prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,dots,f_n$.
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
$endgroup$
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
add a comment |
$begingroup$
Denote by $pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $mathbb{R}^n$, then the operators $Y_i$ defined by
begin{align*}
D(Y_k)&=left{xiinmathscr{H} : int pi_k(theta)^2,dlangle E(theta)xi,xirangle<inftyright},\
Y_kxi&=int pi_k(theta),dE(theta) xi
end{align*}
are self-adjoint with spectral measure $E_k=(pi_k)_#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $fmapsto int f,dE$ is a $ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $mathscr{H}$: For simple functions $f=sum_j a_j 1_{A_j}$ and $g=sum_k b_k 1_{B_k}$ one has
$$
int fg,dE=int sum_{j,k}a_j b_k 1_{A_jcap B_k},dE=sum_{j,k}a_j b_k E(A_j)E( B_k)=left(sum_j a_j E(A_j)right)left(sum_k b_k E(B_k)right)=left(int f,dEright)left(int g,dEright).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
int prod_{k=1}^n f_k(theta_k),dE(theta)=prod_{k=1}^n int f_k(theta_k),dE(theta)=prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,dots,f_n$.
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
$endgroup$
Denote by $pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $mathbb{R}^n$, then the operators $Y_i$ defined by
begin{align*}
D(Y_k)&=left{xiinmathscr{H} : int pi_k(theta)^2,dlangle E(theta)xi,xirangle<inftyright},\
Y_kxi&=int pi_k(theta),dE(theta) xi
end{align*}
are self-adjoint with spectral measure $E_k=(pi_k)_#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $fmapsto int f,dE$ is a $ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $mathscr{H}$: For simple functions $f=sum_j a_j 1_{A_j}$ and $g=sum_k b_k 1_{B_k}$ one has
$$
int fg,dE=int sum_{j,k}a_j b_k 1_{A_jcap B_k},dE=sum_{j,k}a_j b_k E(A_j)E( B_k)=left(sum_j a_j E(A_j)right)left(sum_k b_k E(B_k)right)=left(int f,dEright)left(int g,dEright).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
int prod_{k=1}^n f_k(theta_k),dE(theta)=prod_{k=1}^n int f_k(theta_k),dE(theta)=prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,dots,f_n$.
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
answered Jan 28 at 16:13
MaoWaoMaoWao
3,933618
3,933618
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
add a comment |
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Thanks @MaoYao. So the whole point is: continuing what I was doing, the next step would be to invoke the result you've shown that $fmapsto int f dE$ is a $ast$-homomorphism to commute the product and the integral and write $$T_g=prod int exp(itheta_kg^k) dE(theta)$$ Next, we see that $theta_k = pi_k(theta)$ and hence if $f_k(xi)=exp(ixi g^k)$ we have $$T_g = prod int f_kcirc pi_k (theta)dE(theta).$$ Finaly by definition of the pushforward measure this is $$T_g=prod int f_k(theta_k)dE_k(theta_k).$$
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
$begingroup$
Now, $E_k$ is a spectra measure in its own right, and setting $$Y_k = int theta_k dE_k(theta_k)=int pi_k(theta)dE(theta)$$ it follows that $$T_g = prod_k f_k(Y_k) = prod_k exp(ix^kY_k).$$ Is that the full argument? Thanks a lot again!
$endgroup$
– user1620696
Jan 28 at 17:31
add a comment |
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