Mixing
Prove that $|mathbf{T}^n|^2=sum_{|alpha|=n}frac{n!}{alpha!}|mathbf{T}^{alpha}|^2.$
$begingroup$
Let $E$ be a complex Hilbert space and $mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For ${bf A} = (A_1,...,A_d) in mathcal{L}(E)^d$, the norm of ${bf A}$ is given by
$$|{bf A}|^2=sum_{k=1}^d|A_k|^2.$$
For ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ and ${bf S}=(S_1,cdots,S_m)in mathcal{L}(E)^m$, we set
$$mathbf{T}mathbf{S}:=(T_1S_1,cdots,T_1S_m,T_2S_1,cdots,T_2S_m,cdots,T_dS_1,cdots,T_dS_m).$$
Let $mathbf{T}^2=mathbf{T}mathbf{T}$ and we define by induction $mathbf{T}^{n+1}=mathbf{T}mathbf{T}^n$ for $nin mathbb{N}^*$.
Let $nin mathbb{N}^*$ and ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,jin 1,cdots,d$. I want to prove that
$$|mathbf{T}^n|^2=sum_{|alpha|=n}frac{n!}{alpha!}|mathbf{T}^{alpha}|^2.$$
Note that for $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, we write $alpha!: =alpha_1!cdotsalpha_d!,;|alpha|:=displaystylesum_{j=1}^d|alpha_j|$ and $mathbf{T}^alpha:=T_1^{alpha_1} cdots T_d^{alpha_d}$.
multinomial-coefficients
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
$endgroup$
add a comment |
$begingroup$
Let $E$ be a complex Hilbert space and $mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For ${bf A} = (A_1,...,A_d) in mathcal{L}(E)^d$, the norm of ${bf A}$ is given by
$$|{bf A}|^2=sum_{k=1}^d|A_k|^2.$$
For ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ and ${bf S}=(S_1,cdots,S_m)in mathcal{L}(E)^m$, we set
$$mathbf{T}mathbf{S}:=(T_1S_1,cdots,T_1S_m,T_2S_1,cdots,T_2S_m,cdots,T_dS_1,cdots,T_dS_m).$$
Let $mathbf{T}^2=mathbf{T}mathbf{T}$ and we define by induction $mathbf{T}^{n+1}=mathbf{T}mathbf{T}^n$ for $nin mathbb{N}^*$.
Let $nin mathbb{N}^*$ and ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,jin 1,cdots,d$. I want to prove that
$$|mathbf{T}^n|^2=sum_{|alpha|=n}frac{n!}{alpha!}|mathbf{T}^{alpha}|^2.$$
Note that for $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, we write $alpha!: =alpha_1!cdotsalpha_d!,;|alpha|:=displaystylesum_{j=1}^d|alpha_j|$ and $mathbf{T}^alpha:=T_1^{alpha_1} cdots T_d^{alpha_d}$.
multinomial-coefficients
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
$endgroup$
$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
$begingroup$
Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
$begingroup$
@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19
add a comment |
$begingroup$
Let $E$ be a complex Hilbert space and $mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For ${bf A} = (A_1,...,A_d) in mathcal{L}(E)^d$, the norm of ${bf A}$ is given by
$$|{bf A}|^2=sum_{k=1}^d|A_k|^2.$$
For ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ and ${bf S}=(S_1,cdots,S_m)in mathcal{L}(E)^m$, we set
$$mathbf{T}mathbf{S}:=(T_1S_1,cdots,T_1S_m,T_2S_1,cdots,T_2S_m,cdots,T_dS_1,cdots,T_dS_m).$$
Let $mathbf{T}^2=mathbf{T}mathbf{T}$ and we define by induction $mathbf{T}^{n+1}=mathbf{T}mathbf{T}^n$ for $nin mathbb{N}^*$.
Let $nin mathbb{N}^*$ and ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,jin 1,cdots,d$. I want to prove that
$$|mathbf{T}^n|^2=sum_{|alpha|=n}frac{n!}{alpha!}|mathbf{T}^{alpha}|^2.$$
Note that for $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, we write $alpha!: =alpha_1!cdotsalpha_d!,;|alpha|:=displaystylesum_{j=1}^d|alpha_j|$ and $mathbf{T}^alpha:=T_1^{alpha_1} cdots T_d^{alpha_d}$.
multinomial-coefficients
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
$endgroup$
Let $E$ be a complex Hilbert space and $mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For ${bf A} = (A_1,...,A_d) in mathcal{L}(E)^d$, the norm of ${bf A}$ is given by
$$|{bf A}|^2=sum_{k=1}^d|A_k|^2.$$
For ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ and ${bf S}=(S_1,cdots,S_m)in mathcal{L}(E)^m$, we set
$$mathbf{T}mathbf{S}:=(T_1S_1,cdots,T_1S_m,T_2S_1,cdots,T_2S_m,cdots,T_dS_1,cdots,T_dS_m).$$
Let $mathbf{T}^2=mathbf{T}mathbf{T}$ and we define by induction $mathbf{T}^{n+1}=mathbf{T}mathbf{T}^n$ for $nin mathbb{N}^*$.
Let $nin mathbb{N}^*$ and ${bf T}=(T_1,...,T_d) in mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,jin 1,cdots,d$. I want to prove that
$$|mathbf{T}^n|^2=sum_{|alpha|=n}frac{n!}{alpha!}|mathbf{T}^{alpha}|^2.$$
Note that for $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, we write $alpha!: =alpha_1!cdotsalpha_d!,;|alpha|:=displaystylesum_{j=1}^d|alpha_j|$ and $mathbf{T}^alpha:=T_1^{alpha_1} cdots T_d^{alpha_d}$.
multinomial-coefficients
multinomial-coefficients
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
edited Jan 19 at 6:00
Schüler
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
asked Jan 17 at 11:29
SchülerSchüler
1,5391421
1,5391421
$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
$begingroup$
Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
$begingroup$
@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19
add a comment |
$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
$begingroup$
Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
$begingroup$
@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19
$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
$begingroup$
Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
$begingroup$
Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
$begingroup$
@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19
$begingroup$
@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reason why there is a factor $frac{n!}{alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 $$
where $c_{alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,dots,T_d)$ which correspond to the multi-index $alpha$.
begin{align*}c_{alpha}&=#left{(j_1,dots,j_n):T_{j_1}dots T_{j_n}=T^{alpha},,j_kin left{0,dots,dright}right} = \
&=#left{(j_1,dots,j_n):(#left{j_k=1right}=alpha_1 land dots land #left{j_k=dright}=alpha_d)right}\
end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $alpha$ in the following way
$$underbrace{1,dots, 1}_{alpha_1text{ times }},underbrace{2,dots, 2}_{alpha_2text{ times }},dots, underbrace{d,dots, d}_{alpha_dtext{ times }} $$
Notice that the total amount of numbers written in the above line is $alpha_1+dots+alpha_d=n$.
Then $c_{alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{alpha}=frac{n!}{alpha_1!cdot dots cdot alpha_d!}=frac{n!}{alpha!}$$
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
$endgroup$
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
|
show 8 more comments
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1 Answer
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$begingroup$
The reason why there is a factor $frac{n!}{alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 $$
where $c_{alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,dots,T_d)$ which correspond to the multi-index $alpha$.
begin{align*}c_{alpha}&=#left{(j_1,dots,j_n):T_{j_1}dots T_{j_n}=T^{alpha},,j_kin left{0,dots,dright}right} = \
&=#left{(j_1,dots,j_n):(#left{j_k=1right}=alpha_1 land dots land #left{j_k=dright}=alpha_d)right}\
end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $alpha$ in the following way
$$underbrace{1,dots, 1}_{alpha_1text{ times }},underbrace{2,dots, 2}_{alpha_2text{ times }},dots, underbrace{d,dots, d}_{alpha_dtext{ times }} $$
Notice that the total amount of numbers written in the above line is $alpha_1+dots+alpha_d=n$.
Then $c_{alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{alpha}=frac{n!}{alpha_1!cdot dots cdot alpha_d!}=frac{n!}{alpha!}$$
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
$endgroup$
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
|
show 8 more comments
$begingroup$
The reason why there is a factor $frac{n!}{alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 $$
where $c_{alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,dots,T_d)$ which correspond to the multi-index $alpha$.
begin{align*}c_{alpha}&=#left{(j_1,dots,j_n):T_{j_1}dots T_{j_n}=T^{alpha},,j_kin left{0,dots,dright}right} = \
&=#left{(j_1,dots,j_n):(#left{j_k=1right}=alpha_1 land dots land #left{j_k=dright}=alpha_d)right}\
end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $alpha$ in the following way
$$underbrace{1,dots, 1}_{alpha_1text{ times }},underbrace{2,dots, 2}_{alpha_2text{ times }},dots, underbrace{d,dots, d}_{alpha_dtext{ times }} $$
Notice that the total amount of numbers written in the above line is $alpha_1+dots+alpha_d=n$.
Then $c_{alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{alpha}=frac{n!}{alpha_1!cdot dots cdot alpha_d!}=frac{n!}{alpha!}$$
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
$endgroup$
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
|
show 8 more comments
$begingroup$
The reason why there is a factor $frac{n!}{alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 $$
where $c_{alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,dots,T_d)$ which correspond to the multi-index $alpha$.
begin{align*}c_{alpha}&=#left{(j_1,dots,j_n):T_{j_1}dots T_{j_n}=T^{alpha},,j_kin left{0,dots,dright}right} = \
&=#left{(j_1,dots,j_n):(#left{j_k=1right}=alpha_1 land dots land #left{j_k=dright}=alpha_d)right}\
end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $alpha$ in the following way
$$underbrace{1,dots, 1}_{alpha_1text{ times }},underbrace{2,dots, 2}_{alpha_2text{ times }},dots, underbrace{d,dots, d}_{alpha_dtext{ times }} $$
Notice that the total amount of numbers written in the above line is $alpha_1+dots+alpha_d=n$.
Then $c_{alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{alpha}=frac{n!}{alpha_1!cdot dots cdot alpha_d!}=frac{n!}{alpha!}$$
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
$endgroup$
The reason why there is a factor $frac{n!}{alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 $$
where $c_{alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,dots,T_d)$ which correspond to the multi-index $alpha$.
begin{align*}c_{alpha}&=#left{(j_1,dots,j_n):T_{j_1}dots T_{j_n}=T^{alpha},,j_kin left{0,dots,dright}right} = \
&=#left{(j_1,dots,j_n):(#left{j_k=1right}=alpha_1 land dots land #left{j_k=dright}=alpha_d)right}\
end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $alpha$ in the following way
$$underbrace{1,dots, 1}_{alpha_1text{ times }},underbrace{2,dots, 2}_{alpha_2text{ times }},dots, underbrace{d,dots, d}_{alpha_dtext{ times }} $$
Notice that the total amount of numbers written in the above line is $alpha_1+dots+alpha_d=n$.
Then $c_{alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{alpha}=frac{n!}{alpha_1!cdot dots cdot alpha_d!}=frac{n!}{alpha!}$$
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
edited Jan 17 at 14:47
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
answered Jan 17 at 13:37
Lorenzo QuarisaLorenzo Quarisa
3,630623
3,630623
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For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
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– Schüler
Jan 17 at 14:15
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Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
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– Schüler
Jan 17 at 14:15
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1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
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– Lorenzo Quarisa
Jan 17 at 14:22
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You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
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– Schüler
Jan 17 at 14:41
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
|
show 8 more comments
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
For every $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$, I think that $mathbf{T}^alpha$ is always defined to be $T_1^{alpha_1} cdots T_d^{alpha_d}$. I don't understand why you write ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.''.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
Also I don't understand why $$|T^n|^2=sum_{|alpha|=n}c_{alpha}|T^{alpha}|^2 ?$$ Thank you.
$endgroup$
– Schüler
Jan 17 at 14:15
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
1. because otherwise how could you express e.g. the component $T_2T_1$ (if $d=2$)? In the notation $T_1^{alpha_1}dots T_d^{alpha_d}$ the $T_1$ component always comes first.\ 2. This follows from the definition of the norm of an operator with multiple components, as all the components of $T_n$ are in the form $T^{alpha}$ with $|alpha|=n$ (again assuming that $T_iT_j=T_jT_i$).
$endgroup$
– Lorenzo Quarisa
Jan 17 at 14:22
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
$begingroup$
You mean ''$T^{alpha}$ makes sense precisely because of the assumption that $T_iT_j=T_jT_i$ for all $i,j$.'' when $alpha = (alpha_1,cdots,alpha_d) in mathbb{N}^d$ which is defined as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ do you agree with me?
$endgroup$
– Schüler
Jan 17 at 14:41
1
1
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
$begingroup$
I think here '' which correspond to the multi-index $alpha$, '' $alpha$ is not arbitrary but it is given as $$alpha_k=#{jin{1,cdots,n},;;i_j=k},$$ for all $k=1,cdots,d$.
$endgroup$
– Schüler
Jan 17 at 14:57
|
show 8 more comments
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$begingroup$
have you tried induction in $n$ or $d$?
$endgroup$
– supinf
Jan 17 at 11:58
$begingroup$
@supinf Yes I try induction on $n$ but from $n$ to $n+1$ it is not evident.
$endgroup$
– Schüler
Jan 17 at 12:03
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Also, this question is the same as your old question math.stackexchange.com/questions/3050154/… , except that you are now squaring all monomials (but that is tantamount to squaring all variables, i.e., setting $X_i = T_i^2$).
$endgroup$
– darij grinberg
Jan 17 at 14:54
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@Schüler: Yes, it is true then, more or less by the definition.
$endgroup$
– darij grinberg
Jan 17 at 22:19