When is the set of tensors of rank at most $r$ closed?
Consider the vector space $V=A_1otimes A_2otimesdotsotimes A_k$ where $kgeq 2$ and let $sigma_r$ be the set of tensors in $V$ of rank at most $r$. It seems clear that $sigma_r$ is Zariski (Euclidean?) closed in the following cases :
1 - $k=2$ and $r$ is anything
2 - $r=1$ and $k$ is anything
3 - $sigma_r=V$
Is this list complete? Can I change the third case as $r=prod_{i=1}^k dim A_i$ and if not with some other integral condition on $r$?
algebraic-geometry multilinear-algebra
asked Sep 13 '18 at 14:32
Levent
2,688925
|
show 1 more comment
Consider the vector space $V=A_1otimes A_2otimesdotsotimes A_k$ where $kgeq 2$ and let $sigma_r$ be the set of tensors in $V$ of rank at most $r$. It seems clear that $sigma_r$ is Zariski (Euclidean?) closed in the following cases :
1 - $k=2$ and $r$ is anything
2 - $r=1$ and $k$ is anything
3 - $sigma_r=V$
Is this list complete? Can I change the third case as $r=prod_{i=1}^k dim A_i$ and if not with some other integral condition on $r$?
algebraic-geometry multilinear-algebra
asked Sep 13 '18 at 14:32
Levent
2,688925
What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
3
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
1
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
1
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48
|
show 1 more comment
Consider the vector space $V=A_1otimes A_2otimesdotsotimes A_k$ where $kgeq 2$ and let $sigma_r$ be the set of tensors in $V$ of rank at most $r$. It seems clear that $sigma_r$ is Zariski (Euclidean?) closed in the following cases :
1 - $k=2$ and $r$ is anything
2 - $r=1$ and $k$ is anything
3 - $sigma_r=V$
Is this list complete? Can I change the third case as $r=prod_{i=1}^k dim A_i$ and if not with some other integral condition on $r$?
algebraic-geometry multilinear-algebra
asked Sep 13 '18 at 14:32
Levent
2,688925
Consider the vector space $V=A_1otimes A_2otimesdotsotimes A_k$ where $kgeq 2$ and let $sigma_r$ be the set of tensors in $V$ of rank at most $r$. It seems clear that $sigma_r$ is Zariski (Euclidean?) closed in the following cases :
1 - $k=2$ and $r$ is anything
2 - $r=1$ and $k$ is anything
3 - $sigma_r=V$
Is this list complete? Can I change the third case as $r=prod_{i=1}^k dim A_i$ and if not with some other integral condition on $r$?
algebraic-geometry multilinear-algebra
algebraic-geometry multilinear-algebra
asked Sep 13 '18 at 14:32
Levent
2,688925
asked Sep 13 '18 at 14:32
Levent
2,688925
edited Dec 22 '18 at 15:56
asked Sep 13 '18 at 14:32
Levent
2,688925
asked Sep 13 '18 at 14:32
Levent
2,688925
asked Sep 13 '18 at 14:32
Levent
2,688925
2,688925
What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
3
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
1
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
1
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48
|
show 1 more comment
What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
3
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
1
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
1
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48
What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
3
3
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
1
1
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
1
1
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48
|
show 1 more comment
1 Answer
1
active
oldest
votes
- First of all, the maximal possible rank of degree $k$ tensors is bounded from above by
$$
R=prod_{i=1}^{k-1} dim(A_i).
$$
Here and below the indices are arranged so that
$$2le n_1= dim(A_1)le n_2= dim(A_2)le ...le n_k= dim(A_k).$$
See for instance
M.V. Catalisano, A.V. Geramita, A. Gimigliano,
"Ranks of tensors, secant varieties of Segre varieties and fat points",
Linear Algebra Appl. 355 (2002) 263-285.
This bound is better than the one that you wrote in the last part of your question. In particular, $sigma_R$ equals the entire tensor product
$$
V= A_1otimes ... otimes A_k,
$$
which is, of course, Zariski closed.
- There are some Zariski closed examples besides the ones that you listed. For instance (I am working over ${mathbb C}$),
consider triples
$$
(n_1,n_2,n_3)= (2, 2, n), nge 4.
$$
Then in $V={mathbb C}^2 otimes {mathbb C}^2 otimes {mathbb C}^n$,
all tensors of rank =4 also have the border rank =4, see Table 10.3.1 on page 253 of
J.Landsberg, "Tensors: Geometry and Applications", AMS, 2011.
This means that the set $sigma_3subset V$ of tensors of rank $le 3$ is closed. (In contrast, $sigma_2$ is not closed.)
At the same time, since $nge 4$, $V=sigma_4ne sigma_3$.
I very much doubt that a complete list of tuples $(n_1,...,n_k,r)$ for which $sigma_r$ is Zariski closed is known. Among general results on lack of closedness in the classical topology the following is known, see
V. De Silva, L.-H. Lim, "Tensor rank and the ill-posedness of the best low-rank approximation problem," SIAM J. Matrix Anal. Appl. 30 (2008) 1084-1127.
Theorem. Suppose $kge 3$ and
$$
2le rle min(n_1,...,n_k).
$$
Then $sigma_r$ is not closed in the classical topology. (Hence, it is not closed in the Zariski topology either.)
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
add a comment |
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1 Answer
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- First of all, the maximal possible rank of degree $k$ tensors is bounded from above by
$$
R=prod_{i=1}^{k-1} dim(A_i).
$$
Here and below the indices are arranged so that
$$2le n_1= dim(A_1)le n_2= dim(A_2)le ...le n_k= dim(A_k).$$
See for instance
M.V. Catalisano, A.V. Geramita, A. Gimigliano,
"Ranks of tensors, secant varieties of Segre varieties and fat points",
Linear Algebra Appl. 355 (2002) 263-285.
This bound is better than the one that you wrote in the last part of your question. In particular, $sigma_R$ equals the entire tensor product
$$
V= A_1otimes ... otimes A_k,
$$
which is, of course, Zariski closed.
- There are some Zariski closed examples besides the ones that you listed. For instance (I am working over ${mathbb C}$),
consider triples
$$
(n_1,n_2,n_3)= (2, 2, n), nge 4.
$$
Then in $V={mathbb C}^2 otimes {mathbb C}^2 otimes {mathbb C}^n$,
all tensors of rank =4 also have the border rank =4, see Table 10.3.1 on page 253 of
J.Landsberg, "Tensors: Geometry and Applications", AMS, 2011.
This means that the set $sigma_3subset V$ of tensors of rank $le 3$ is closed. (In contrast, $sigma_2$ is not closed.)
At the same time, since $nge 4$, $V=sigma_4ne sigma_3$.
I very much doubt that a complete list of tuples $(n_1,...,n_k,r)$ for which $sigma_r$ is Zariski closed is known. Among general results on lack of closedness in the classical topology the following is known, see
V. De Silva, L.-H. Lim, "Tensor rank and the ill-posedness of the best low-rank approximation problem," SIAM J. Matrix Anal. Appl. 30 (2008) 1084-1127.
Theorem. Suppose $kge 3$ and
$$
2le rle min(n_1,...,n_k).
$$
Then $sigma_r$ is not closed in the classical topology. (Hence, it is not closed in the Zariski topology either.)
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
add a comment |
- First of all, the maximal possible rank of degree $k$ tensors is bounded from above by
$$
R=prod_{i=1}^{k-1} dim(A_i).
$$
Here and below the indices are arranged so that
$$2le n_1= dim(A_1)le n_2= dim(A_2)le ...le n_k= dim(A_k).$$
See for instance
M.V. Catalisano, A.V. Geramita, A. Gimigliano,
"Ranks of tensors, secant varieties of Segre varieties and fat points",
Linear Algebra Appl. 355 (2002) 263-285.
This bound is better than the one that you wrote in the last part of your question. In particular, $sigma_R$ equals the entire tensor product
$$
V= A_1otimes ... otimes A_k,
$$
which is, of course, Zariski closed.
- There are some Zariski closed examples besides the ones that you listed. For instance (I am working over ${mathbb C}$),
consider triples
$$
(n_1,n_2,n_3)= (2, 2, n), nge 4.
$$
Then in $V={mathbb C}^2 otimes {mathbb C}^2 otimes {mathbb C}^n$,
all tensors of rank =4 also have the border rank =4, see Table 10.3.1 on page 253 of
J.Landsberg, "Tensors: Geometry and Applications", AMS, 2011.
This means that the set $sigma_3subset V$ of tensors of rank $le 3$ is closed. (In contrast, $sigma_2$ is not closed.)
At the same time, since $nge 4$, $V=sigma_4ne sigma_3$.
I very much doubt that a complete list of tuples $(n_1,...,n_k,r)$ for which $sigma_r$ is Zariski closed is known. Among general results on lack of closedness in the classical topology the following is known, see
V. De Silva, L.-H. Lim, "Tensor rank and the ill-posedness of the best low-rank approximation problem," SIAM J. Matrix Anal. Appl. 30 (2008) 1084-1127.
Theorem. Suppose $kge 3$ and
$$
2le rle min(n_1,...,n_k).
$$
Then $sigma_r$ is not closed in the classical topology. (Hence, it is not closed in the Zariski topology either.)
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
add a comment |
- First of all, the maximal possible rank of degree $k$ tensors is bounded from above by
$$
R=prod_{i=1}^{k-1} dim(A_i).
$$
Here and below the indices are arranged so that
$$2le n_1= dim(A_1)le n_2= dim(A_2)le ...le n_k= dim(A_k).$$
See for instance
M.V. Catalisano, A.V. Geramita, A. Gimigliano,
"Ranks of tensors, secant varieties of Segre varieties and fat points",
Linear Algebra Appl. 355 (2002) 263-285.
This bound is better than the one that you wrote in the last part of your question. In particular, $sigma_R$ equals the entire tensor product
$$
V= A_1otimes ... otimes A_k,
$$
which is, of course, Zariski closed.
- There are some Zariski closed examples besides the ones that you listed. For instance (I am working over ${mathbb C}$),
consider triples
$$
(n_1,n_2,n_3)= (2, 2, n), nge 4.
$$
Then in $V={mathbb C}^2 otimes {mathbb C}^2 otimes {mathbb C}^n$,
all tensors of rank =4 also have the border rank =4, see Table 10.3.1 on page 253 of
J.Landsberg, "Tensors: Geometry and Applications", AMS, 2011.
This means that the set $sigma_3subset V$ of tensors of rank $le 3$ is closed. (In contrast, $sigma_2$ is not closed.)
At the same time, since $nge 4$, $V=sigma_4ne sigma_3$.
I very much doubt that a complete list of tuples $(n_1,...,n_k,r)$ for which $sigma_r$ is Zariski closed is known. Among general results on lack of closedness in the classical topology the following is known, see
V. De Silva, L.-H. Lim, "Tensor rank and the ill-posedness of the best low-rank approximation problem," SIAM J. Matrix Anal. Appl. 30 (2008) 1084-1127.
Theorem. Suppose $kge 3$ and
$$
2le rle min(n_1,...,n_k).
$$
Then $sigma_r$ is not closed in the classical topology. (Hence, it is not closed in the Zariski topology either.)
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
- First of all, the maximal possible rank of degree $k$ tensors is bounded from above by
$$
R=prod_{i=1}^{k-1} dim(A_i).
$$
Here and below the indices are arranged so that
$$2le n_1= dim(A_1)le n_2= dim(A_2)le ...le n_k= dim(A_k).$$
See for instance
M.V. Catalisano, A.V. Geramita, A. Gimigliano,
"Ranks of tensors, secant varieties of Segre varieties and fat points",
Linear Algebra Appl. 355 (2002) 263-285.
This bound is better than the one that you wrote in the last part of your question. In particular, $sigma_R$ equals the entire tensor product
$$
V= A_1otimes ... otimes A_k,
$$
which is, of course, Zariski closed.
- There are some Zariski closed examples besides the ones that you listed. For instance (I am working over ${mathbb C}$),
consider triples
$$
(n_1,n_2,n_3)= (2, 2, n), nge 4.
$$
Then in $V={mathbb C}^2 otimes {mathbb C}^2 otimes {mathbb C}^n$,
all tensors of rank =4 also have the border rank =4, see Table 10.3.1 on page 253 of
J.Landsberg, "Tensors: Geometry and Applications", AMS, 2011.
This means that the set $sigma_3subset V$ of tensors of rank $le 3$ is closed. (In contrast, $sigma_2$ is not closed.)
At the same time, since $nge 4$, $V=sigma_4ne sigma_3$.
I very much doubt that a complete list of tuples $(n_1,...,n_k,r)$ for which $sigma_r$ is Zariski closed is known. Among general results on lack of closedness in the classical topology the following is known, see
V. De Silva, L.-H. Lim, "Tensor rank and the ill-posedness of the best low-rank approximation problem," SIAM J. Matrix Anal. Appl. 30 (2008) 1084-1127.
Theorem. Suppose $kge 3$ and
$$
2le rle min(n_1,...,n_k).
$$
Then $sigma_r$ is not closed in the classical topology. (Hence, it is not closed in the Zariski topology either.)
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
edited Dec 26 '18 at 23:27
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
answered Dec 26 '18 at 23:03
Moishe Cohen
45.8k342104
45.8k342104
add a comment |
add a comment |
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What is your definition for rank of a tensor?
– KReiser
Sep 13 '18 at 18:18
3
A rank one tensor is a tensor $Tin V$ of the form $T=a_1otimes a_2otimesdotsotimes a_k$. The rank of a tensor $T$ is the smallest number $r$ such that $T$ can be written as the sum of $r$ rank one tensors.
– Levent
Sep 13 '18 at 20:21
1
If $V$ has the structure of a finite dimensional (real hausdorff) topological vector space, then: yes. Since you mention the Zariski topology (and euclidean?) I am not sure what your setting is.
– s.harp
Nov 21 '18 at 9:58
1
Above, each $A_i$ is a finite dimensional complex vector space hence $V$ admits the Zariski topology.
– Levent
Nov 21 '18 at 10:05
You have to specify which topology you are working with: Zariski topology or Euclidean topology. I think the answer to your question depends on this.
– Moishe Cohen
Dec 24 '18 at 15:48