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Contraction of Tensors is Independent of the Choice of Basis
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Definition: Let $ T:(V^*)^k times V^l rightarrow mathbbR$ be a tensor of type $ (k,l)$. Let $ v_1,...,v_n$ be a basis of $V$ and $ v^1^*,...,v^n^*$ be the corresponding dual basis. The contraction of $ T$ with respect to the $ i$th (dual vector) and $j$th slot (which is a tensor of type $(k-1,l-1)$) is given by
$displaystyle CT=sum_k=1^n T(bullet,...,v^k^*,...,bullet; bullet,...,v_k,...,bullet)$
where the substitutions are done at the $i$th dual vector slot and the $j$th vector slot.
Theorem: The contraction of a tensor is well-defined, i.e. it is independent of the choice of the basis of $V$.
Proof: Recall that if $v in V$, $w in V^*$, then we have
$$displaystyle v=sum_k=1^n v^k^*(v)v_k$$
$$displaystyle w=sum_k=1^n w(v_k)v^k^*$$
Let $v_1',...,v_n'$ be another basis of $V$ and $v^1^*',...,v^n^*'$ be the corresponding dual basis. Then we have
$$displaystyle v_i'=sum_k=1^n v^k^*(v_i')v_k$$
$$displaystyle v^i^*'=sum_k=1^nv^i^*'(v_k)v^k^*$$
Hence, we get
$$beginaligned displaystyle sum_k=1^n T(bullet,...,v^k^*',...,bullet; bullet,...,v_k',...,bullet) &=sum_k=1^n T(bullet,...,sum_j=1^nv^k^*'(v_j)v^j^*,...,bullet; bullet,...,sum_p=1^n v^p^*(v_k')v_p,...,bullet)\&= sum_k=1^nsum_p=1^nsum_j=1^n v^k^*'(v_j)v^p^*(v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)\&=sum_k=1^nsum_p=1^nsum_j=1^n v^p^*(v^k^*'(v_j)v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)endaligned$$
I have no idea how to proceed. Can anyone give me some suggestions? Thanks.
linear-algebra differential-geometry tensors multilinear-algebra
edited Aug 2 at 14:29
Sou
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asked Aug 2 at 14:05
Jerry
374211
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up vote
2
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Definition: Let $ T:(V^*)^k times V^l rightarrow mathbbR$ be a tensor of type $ (k,l)$. Let $ v_1,...,v_n$ be a basis of $V$ and $ v^1^*,...,v^n^*$ be the corresponding dual basis. The contraction of $ T$ with respect to the $ i$th (dual vector) and $j$th slot (which is a tensor of type $(k-1,l-1)$) is given by
$displaystyle CT=sum_k=1^n T(bullet,...,v^k^*,...,bullet; bullet,...,v_k,...,bullet)$
where the substitutions are done at the $i$th dual vector slot and the $j$th vector slot.
Theorem: The contraction of a tensor is well-defined, i.e. it is independent of the choice of the basis of $V$.
Proof: Recall that if $v in V$, $w in V^*$, then we have
$$displaystyle v=sum_k=1^n v^k^*(v)v_k$$
$$displaystyle w=sum_k=1^n w(v_k)v^k^*$$
Let $v_1',...,v_n'$ be another basis of $V$ and $v^1^*',...,v^n^*'$ be the corresponding dual basis. Then we have
$$displaystyle v_i'=sum_k=1^n v^k^*(v_i')v_k$$
$$displaystyle v^i^*'=sum_k=1^nv^i^*'(v_k)v^k^*$$
Hence, we get
$$beginaligned displaystyle sum_k=1^n T(bullet,...,v^k^*',...,bullet; bullet,...,v_k',...,bullet) &=sum_k=1^n T(bullet,...,sum_j=1^nv^k^*'(v_j)v^j^*,...,bullet; bullet,...,sum_p=1^n v^p^*(v_k')v_p,...,bullet)\&= sum_k=1^nsum_p=1^nsum_j=1^n v^k^*'(v_j)v^p^*(v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)\&=sum_k=1^nsum_p=1^nsum_j=1^n v^p^*(v^k^*'(v_j)v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)endaligned$$
I have no idea how to proceed. Can anyone give me some suggestions? Thanks.
linear-algebra differential-geometry tensors multilinear-algebra
edited Aug 2 at 14:29
Sou
2,6192820
asked Aug 2 at 14:05
Jerry
374211
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Definition: Let $ T:(V^*)^k times V^l rightarrow mathbbR$ be a tensor of type $ (k,l)$. Let $ v_1,...,v_n$ be a basis of $V$ and $ v^1^*,...,v^n^*$ be the corresponding dual basis. The contraction of $ T$ with respect to the $ i$th (dual vector) and $j$th slot (which is a tensor of type $(k-1,l-1)$) is given by
$displaystyle CT=sum_k=1^n T(bullet,...,v^k^*,...,bullet; bullet,...,v_k,...,bullet)$
where the substitutions are done at the $i$th dual vector slot and the $j$th vector slot.
Theorem: The contraction of a tensor is well-defined, i.e. it is independent of the choice of the basis of $V$.
Proof: Recall that if $v in V$, $w in V^*$, then we have
$$displaystyle v=sum_k=1^n v^k^*(v)v_k$$
$$displaystyle w=sum_k=1^n w(v_k)v^k^*$$
Let $v_1',...,v_n'$ be another basis of $V$ and $v^1^*',...,v^n^*'$ be the corresponding dual basis. Then we have
$$displaystyle v_i'=sum_k=1^n v^k^*(v_i')v_k$$
$$displaystyle v^i^*'=sum_k=1^nv^i^*'(v_k)v^k^*$$
Hence, we get
$$beginaligned displaystyle sum_k=1^n T(bullet,...,v^k^*',...,bullet; bullet,...,v_k',...,bullet) &=sum_k=1^n T(bullet,...,sum_j=1^nv^k^*'(v_j)v^j^*,...,bullet; bullet,...,sum_p=1^n v^p^*(v_k')v_p,...,bullet)\&= sum_k=1^nsum_p=1^nsum_j=1^n v^k^*'(v_j)v^p^*(v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)\&=sum_k=1^nsum_p=1^nsum_j=1^n v^p^*(v^k^*'(v_j)v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)endaligned$$
I have no idea how to proceed. Can anyone give me some suggestions? Thanks.
linear-algebra differential-geometry tensors multilinear-algebra
edited Aug 2 at 14:29
Sou
2,6192820
asked Aug 2 at 14:05
Jerry
374211
Definition: Let $ T:(V^*)^k times V^l rightarrow mathbbR$ be a tensor of type $ (k,l)$. Let $ v_1,...,v_n$ be a basis of $V$ and $ v^1^*,...,v^n^*$ be the corresponding dual basis. The contraction of $ T$ with respect to the $ i$th (dual vector) and $j$th slot (which is a tensor of type $(k-1,l-1)$) is given by
$displaystyle CT=sum_k=1^n T(bullet,...,v^k^*,...,bullet; bullet,...,v_k,...,bullet)$
where the substitutions are done at the $i$th dual vector slot and the $j$th vector slot.
Theorem: The contraction of a tensor is well-defined, i.e. it is independent of the choice of the basis of $V$.
Proof: Recall that if $v in V$, $w in V^*$, then we have
$$displaystyle v=sum_k=1^n v^k^*(v)v_k$$
$$displaystyle w=sum_k=1^n w(v_k)v^k^*$$
Let $v_1',...,v_n'$ be another basis of $V$ and $v^1^*',...,v^n^*'$ be the corresponding dual basis. Then we have
$$displaystyle v_i'=sum_k=1^n v^k^*(v_i')v_k$$
$$displaystyle v^i^*'=sum_k=1^nv^i^*'(v_k)v^k^*$$
Hence, we get
$$beginaligned displaystyle sum_k=1^n T(bullet,...,v^k^*',...,bullet; bullet,...,v_k',...,bullet) &=sum_k=1^n T(bullet,...,sum_j=1^nv^k^*'(v_j)v^j^*,...,bullet; bullet,...,sum_p=1^n v^p^*(v_k')v_p,...,bullet)\&= sum_k=1^nsum_p=1^nsum_j=1^n v^k^*'(v_j)v^p^*(v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)\&=sum_k=1^nsum_p=1^nsum_j=1^n v^p^*(v^k^*'(v_j)v_k')T(bullet,...,v^j^*,...,bullet; bullet,...,v_p,...,bullet)endaligned$$
I have no idea how to proceed. Can anyone give me some suggestions? Thanks.
linear-algebra differential-geometry tensors multilinear-algebra
edited Aug 2 at 14:29
Sou
2,6192820
asked Aug 2 at 14:05
Jerry
374211
edited Aug 2 at 14:29
Sou
2,6192820
edited Aug 2 at 14:29
Sou
2,6192820
edited Aug 2 at 14:29
Sou
2,6192820
2,6192820
asked Aug 2 at 14:05
Jerry
374211
asked Aug 2 at 14:05
Jerry
374211
asked Aug 2 at 14:05
Jerry
374211
374211
add a comment |Â
add a comment |Â
2 Answers
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You already wrote everything. In the last line you wrote $v^k^*'(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^p^*(v_j)=delta^p^*_j$ and conclude.
answered Aug 2 at 14:29
Lolman
919612
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
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up vote
1
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I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.
Let's simplify to the case of a (1,1) tensor $ omega otimes v$. Denote contraction by C, then $$C (omega otimes v )= omega (v)=omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.
Another way to see this is that a (1,1) tensor is a linear map $Vrightarrow V$ and contraction is the trace of this linear map.
answered Aug 2 at 14:24
dave
385
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You already wrote everything. In the last line you wrote $v^k^*'(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^p^*(v_j)=delta^p^*_j$ and conclude.
answered Aug 2 at 14:29
Lolman
919612
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
add a comment |Â
up vote
2
down vote
accepted
You already wrote everything. In the last line you wrote $v^k^*'(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^p^*(v_j)=delta^p^*_j$ and conclude.
answered Aug 2 at 14:29
Lolman
919612
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You already wrote everything. In the last line you wrote $v^k^*'(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^p^*(v_j)=delta^p^*_j$ and conclude.
answered Aug 2 at 14:29
Lolman
919612
You already wrote everything. In the last line you wrote $v^k^*'(v_j)v'_k$ but that is exactly $v_j$ as you wrote in the above relations. You substitute it in your equation and get $v^p^*(v_j)=delta^p^*_j$ and conclude.
answered Aug 2 at 14:29
Lolman
919612
answered Aug 2 at 14:29
Lolman
919612
answered Aug 2 at 14:29
Lolman
919612
answered Aug 2 at 14:29
Lolman
919612
919612
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
add a comment |Â
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
Oh I see! I just need to put the outermost summation into the $ v^k^*'v_k'$ right?
– Jerry
Aug 2 at 14:43
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
@Jerry That is exacly what's left to do!
– Lolman
Aug 3 at 2:42
add a comment |Â
up vote
1
down vote
I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.
Let's simplify to the case of a (1,1) tensor $ omega otimes v$. Denote contraction by C, then $$C (omega otimes v )= omega (v)=omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.
Another way to see this is that a (1,1) tensor is a linear map $Vrightarrow V$ and contraction is the trace of this linear map.
answered Aug 2 at 14:24
dave
385
add a comment |Â
up vote
1
down vote
I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.
Let's simplify to the case of a (1,1) tensor $ omega otimes v$. Denote contraction by C, then $$C (omega otimes v )= omega (v)=omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.
Another way to see this is that a (1,1) tensor is a linear map $Vrightarrow V$ and contraction is the trace of this linear map.
answered Aug 2 at 14:24
dave
385
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.
Let's simplify to the case of a (1,1) tensor $ omega otimes v$. Denote contraction by C, then $$C (omega otimes v )= omega (v)=omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.
Another way to see this is that a (1,1) tensor is a linear map $Vrightarrow V$ and contraction is the trace of this linear map.
answered Aug 2 at 14:24
dave
385
I don't have enough reputation to write a comment, so I'll write a short answer. You seem to make the problem rather complicated.
Let's simplify to the case of a (1,1) tensor $ omega otimes v$. Denote contraction by C, then $$C (omega otimes v )= omega (v)=omega ^i v_i.$$ Under a coordinate transformation, i.e. a change of basis, the components of $omega $ and $v$ transform inverse to each other, hence contraction is independent of the basis.
Another way to see this is that a (1,1) tensor is a linear map $Vrightarrow V$ and contraction is the trace of this linear map.
answered Aug 2 at 14:24
dave
385
answered Aug 2 at 14:24
dave
385
answered Aug 2 at 14:24
dave
385
answered Aug 2 at 14:24
dave
385
385
add a comment |Â
add a comment |Â
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