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When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?
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Let $1 leq k < n$ be natural numbers. Given orthonormal vectors $u_1,dots,u_k$ in $bf R^n$, one can always find an additional unit vector $v in bf R^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,dots,u_k$, as the tuple $(u_1,dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,dots,u_k$ that is consistent with a chosen orientation on $bf R^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.
at.algebraic-topology grassmannians orthogonal-matrices
asked 1 hour ago
Terry Tao
56.5k17245335
add a comment |Â
up vote
6
down vote
favorite
Let $1 leq k < n$ be natural numbers. Given orthonormal vectors $u_1,dots,u_k$ in $bf R^n$, one can always find an additional unit vector $v in bf R^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,dots,u_k$, as the tuple $(u_1,dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,dots,u_k$ that is consistent with a chosen orientation on $bf R^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.
at.algebraic-topology grassmannians orthogonal-matrices
asked 1 hour ago
Terry Tao
56.5k17245335
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $1 leq k < n$ be natural numbers. Given orthonormal vectors $u_1,dots,u_k$ in $bf R^n$, one can always find an additional unit vector $v in bf R^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,dots,u_k$, as the tuple $(u_1,dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,dots,u_k$ that is consistent with a chosen orientation on $bf R^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.
at.algebraic-topology grassmannians orthogonal-matrices
asked 1 hour ago
Terry Tao
56.5k17245335
Let $1 leq k < n$ be natural numbers. Given orthonormal vectors $u_1,dots,u_k$ in $bf R^n$, one can always find an additional unit vector $v in bf R^n$ that is orthogonal to the preceding $k$. My question is: under what conditions on $k,n$ is it possible to make $v$ depend continuously on $u_1,dots,u_k$, as the tuple $(u_1,dots,u_k)$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $k=n-1$ then one can just pick the unique unit normal to the span of the $u_1,dots,u_k$ that is consistent with a chosen orientation on $bf R^n$ (i.e., take wedge product and then Hodge dual, or just cross product in the $(k,n)=(2,3)$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $n$ is much larger than $k$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $Gr(k,n)$, though I don't know how to calculate the space of such sections.
at.algebraic-topology grassmannians orthogonal-matrices
at.algebraic-topology grassmannians orthogonal-matrices
asked 1 hour ago
Terry Tao
56.5k17245335
asked 1 hour ago
Terry Tao
56.5k17245335
asked 1 hour ago
Terry Tao
56.5k17245335
asked 1 hour ago
Terry Tao
56.5k17245335
asked 1 hour ago
Terry Tao
56.5k17245335
56.5k17245335
add a comment |Â
add a comment |Â
1 Answer
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up vote
5
down vote
Unless I'm missing something, I think that the harry ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
answered 1 hour ago
Will Brian
8,02423652
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Unless I'm missing something, I think that the harry ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
answered 1 hour ago
Will Brian
8,02423652
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
add a comment |Â
up vote
5
down vote
Unless I'm missing something, I think that the harry ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
answered 1 hour ago
Will Brian
8,02423652
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Unless I'm missing something, I think that the harry ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
answered 1 hour ago
Will Brian
8,02423652
Unless I'm missing something, I think that the harry ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
answered 1 hour ago
Will Brian
8,02423652
answered 1 hour ago
Will Brian
8,02423652
answered 1 hour ago
Will Brian
8,02423652
answered 1 hour ago
Will Brian
8,02423652
8,02423652
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
add a comment |Â
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$.
– R. van Dobben de Bruyn
50 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Maybe I am misunderstanding, but if $u_1$ is a radial vector field, then what does it do at the origin?
– David Hughes
40 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently.
– Terry Tao
29 mins ago
add a comment |Â
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