From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Q: dimension of this set
Date: 21 Aug 1995 05:48:25 GMT
In article <1995Aug16.083801.22159@news.snu.ac.kr>,
PooSung Park wrote:
> M = {(V1, V2, V3)| V1, V2, V3 are linear subspaces of R^8
> of dimensions 1, 3, 5, respectively
> such that V3 contains V2 and V2 contains V1.}
>
> dim M = ?
dim M = 23.
In general, GL(n, R) acts transitively on the set of linear subspaces
of dimension k (k <= n); applying this to various combinations of
k and n, we see M is the orbit of a single triple such as
(R^1, R^3, R^5) under the action of GL(8,R). The stabilizer of any
triple in M is conjugate to the stabilizer of this one triple, which is
the collection of (invertible) 8 x 8 matrices of a certain form: the first
column is non-zero only in the first row; the first three columns are
non-zero only in the first three rows; and the first five columns are
non-zero only in the first five rows. The dimension of the set of these is
1+(2+4)+(2+4+4)+(3+6+6+9)=41. Since the whole of GL(8) is of dimension 64,
the codimension of the stabilizer (and thus the dimension of M) is
64-41=23 (which you could also get as the number of zeros in the stabilizer).
The keyword you need is "flag manifolds".
dave
==============================================================================
From: wiggerma@sci.kun.nl (Mark Wiggerman)
Newsgroups: sci.math
Subject: Re: Q: dimension of this set
Date: 22 Aug 1995 10:58:13 GMT
In <1995Aug16.083801.22159@news.snu.ac.kr> sung@math.snu.ac.kr (PooSung Park) writes:
> M = {(V1, V2, V3)| V1, V2, V3 are linear subspaces of R^8
> of dimensions 1, 3, 5, respectively
> such that V3 contains V2 and V2 contains V1.}
> dim M = ?
Indeed, the dimension is 23. This is an example of a classical flag manifold.
This manifold shows much structure ! It can be give the structure of a
CW-complex, and its fundamental group and cohomology groups can be calculated:
Its fundamental group is the direct product of three cyclic groups of order 2.
Generators can be indicated explicitly on M.
Mark
==============================================================================
Date: Mon, 21 Aug 1995 13:45:46 -0800
From: prezky@apple.com (Michael Press)
To: rusin@washington.math.niu.edu (Dave Rusin)
Subject: Re: Q: dimension of this set
In article <4196n9$ajk@watson.math.niu.edu>, rusin@washington.math.niu.edu (Dave Rusin) wrote:
>In article <1995Aug16.083801.22159@news.snu.ac.kr>,
>PooSung Park wrote:
>> M = {(V1, V2, V3)| V1, V2, V3 are linear subspaces of R^8
>> of dimensions 1, 3, 5, respectively
>> such that V3 contains V2 and V2 contains V1.}
>>
>> dim M = ?
>
>dim M = 23.
>
>In general, GL(n, R) acts transitively on the set of linear subspaces
>of dimension k (k <= n); applying this to various combinations of
>k and n, we see M is the orbit of a single triple such as
>(R^1, R^3, R^5) under the action of GL(8,R). The stabilizer of any
>triple in M is conjugate to the stabilizer of this one triple, which is
>the collection of (invertible) 8 x 8 matrices of a certain form: the first
>column is non-zero only in the first row; the first three columns are
>non-zero only in the first three rows; and the first five columns are
>non-zero only in the first five rows. The dimension of the set of these is
>1+(2+4)+(2+4+4)+(3+6+6+9)=41. Since the whole of GL(8) is of dimension 64,
>the codimension of the stabilizer (and thus the dimension of M) is
>64-41=23 (which you could also get as the number of zeros in the stabilizer).
>
>The keyword you need is "flag manifolds".
>
>dave
Thank you for your illuminating article.
I follow the argument until you start adding numbers to get the dimension of the stabilizer. Since the stabilizer is in SL(8,R) where do you account for the fact there is one less 'degree of freedom' in constructing an element of SL(8,R). Also, will you spell out the your scheme for counting ways of constructing an element of the desired stabilizer?
Thank you,
MIchael Press
--
Michael Press
prezky@apple.com
==============================================================================
Date: Tue, 22 Aug 95 10:00:30 CDT
From: rusin (Dave Rusin)
To: prezky@apple.com
Subject: Re: Q: dimension of this set
You wrote me:
>>In article <1995Aug16.083801.22159@news.snu.ac.kr>,
>>PooSung Park wrote:
>>> M = {(V1, V2, V3)| V1, V2, V3 are linear subspaces of R^8
>>> of dimensions 1, 3, 5, respectively
>>> such that V3 contains V2 and V2 contains V1.}
>>>
>>> dim M = ?
[Rusin speaking:]
>>dim M = 23.
>>
>>In general, GL(n, R) acts transitively on the set of linear subspaces
>>of dimension k (k <= n); applying this to various combinations of
>>k and n, we see M is the orbit of a single triple such as
>>(R^1, R^3, R^5) under the action of GL(8,R). The stabilizer of any
>>triple in M is conjugate to the stabilizer of this one triple, which is
>>the collection of (invertible) 8 x 8 matrices of a certain form: the first
>>column is non-zero only in the first row; the first three columns are
>>non-zero only in the first three rows; and the first five columns are
>>non-zero only in the first five rows. The dimension of the set of these is
>>1+(2+4)+(2+4+4)+(3+6+6+9)=41. Since the whole of GL(8) is of dimension 64,
>>the codimension of the stabilizer (and thus the dimension of M) is
>>64-41=23 (which you could also get as the number of zeros in the stabilizer).
>>
>>The keyword you need is "flag manifolds".
>>
>>dave
[your question:]
>I follow the argument until you start adding numbers to get the
>dimension of the stabilizer. Since the stabilizer is in SL(8,R) where
I disagree; for example every diagonal matrix clearly lies in the
stabilizer, so you have matrices of all determinants in there.
Perhaps there is some confusion on the action: GL(8) permutes these
subspaces by left-multiplication. This action being linear, it carries
subspaces to subspaces. The stabilizer is the set of matrices which
preserves this chain of _subspaces_, not _vectors_. (For example, the
diagonal matrix with 2's on the diagonal clearly fixes no non-zero _vector_,
but preserves every _subspace_.)
>do you account for the fact there is one less 'degree of freedom' in
>constructing an element of SL(8,R). Also, will you spell out the your
>scheme for counting ways of constructing an element of the desired
>stabilizer?
Certainly. Recall that I'm looking for the stabilizer of the particular
chain (R^1, R^3, R^5), by which I mean the subspaces of R^8 whose
last few (7,5, or 3) entries are zero. This stabilizer consists of
matrices M making M V_i = V_i for each of the these three linear
subspaces V_i.
The first condition, M V_1 = V_1, is satisfied iff the one basis element
e_1 of V_1 is carried to a multiple of itself, that is, in the
expansion M e_1 = Sum m_i1 e_i we must have m_1i = 0 for all i>1.
(We also need m_11 <> 0 , but that will then be forced by the invertibility
of M.)
The next condition, M V_2 = V_2, is satisfied iff each of the three
basis elements e_1, e_2, and e_3 of V_2 is carried back into V_2
(Again this really only gives M V_2 to be a _subset_ of V_2, but
the invertibility of M disallows any shrinking of dimensions.) Well,
we already have M e_1 being a multiple of e_1, but we also need
M e_2 = m_12 e_1 + m_22 e_2 + m_32 e_3 for some constants m_i2,
and likewise M e_3. This gives further coefficients of the matrix
M which must be zero.
Finally the condition M V_3 = V_3 is similarly expressed as the condition
that M have only zeros in the first five columns, below the first five
rows.
Thus M must have a characteristic "staircase" upperdiagonal form, the
diagonal chunks having dimensions 1, 3-1=2, 5-3=2, and 8-5=3. You run into
this form all the time in this kind of discussion. It shows up a lot in
texts on homogeneous spaces, Lie Group actions, projective geometry, and
so on.
Let me know if this is still unclear.
dave
==============================================================================
From: Michael Press
Subject: Re: Q: dimension of this set
To: rusin@math.niu.edu (Dave Rusin)
Date: Tue, 22 Aug 95 13:17:01 PDT
> You wrote me:
> >>In article <1995Aug16.083801.22159@news.snu.ac.kr>,
> >>PooSung Park wrote:
> >>> M = {(V1, V2, V3)| V1, V2, V3 are linear subspaces of R^8
> >>> of dimensions 1, 3, 5, respectively
> >>> such that V3 contains V2 and V2 contains V1.}
> >>>
> >>> dim M = ?
> [Rusin speaking:]
> >>dim M = 23.
> >>
> >>In general, GL(n, R) acts transitively on the set of linear subspaces
> >>of dimension k (k <= n); applying this to various combinations of
> >>k and n, we see M is the orbit of a single triple such as
> >>(R^1, R^3, R^5) under the action of GL(8,R). The stabilizer of any
> >>triple in M is conjugate to the stabilizer of this one triple, which is
> >>the collection of (invertible) 8 x 8 matrices of a certain form: the first
> >>column is non-zero only in the first row; the first three columns are
> >>non-zero only in the first three rows; and the first five columns are
> >>non-zero only in the first five rows. The dimension of the set of these is
> >>1+(2+4)+(2+4+4)+(3+6+6+9)=41. Since the whole of GL(8) is of dimension 64,
> >>the codimension of the stabilizer (and thus the dimension of M) is
> >>64-41=23 (which you could also get as the number of zeros in the stabilizer).
> >>
> >>The keyword you need is "flag manifolds".
> >>
> >>dave
>
> [your question:]
> >I follow the argument until you start adding numbers to get the
> >dimension of the stabilizer. Since the stabilizer is in SL(8,R) where
>
> I disagree; for example every diagonal matrix clearly lies in the
> stabilizer, so you have matrices of all determinants in there.
Yes, I recognized that the stabilizer needs to be invertible, and then
immediately confused that with the property of having unit determinant.
>
> Perhaps there is some confusion on the action: GL(8) permutes these
> subspaces by left-multiplication. This action being linear, it carries
> subspaces to subspaces. The stabilizer is the set of matrices which
> preserves this chain of _subspaces_, not _vectors_. (For example, the
> diagonal matrix with 2's on the diagonal clearly fixes no non-zero _vector_,
> but preserves every _subspace_.)
>
> >do you account for the fact there is one less 'degree of freedom' in
> >constructing an element of SL(8,R). Also, will you spell out the your
> >scheme for counting ways of constructing an element of the desired
> >stabilizer?
>
> Certainly. Recall that I'm looking for the stabilizer of the particular
> chain (R^1, R^3, R^5), by which I mean the subspaces of R^8 whose
> last few (7,5, or 3) entries are zero. This stabilizer consists of
> matrices M making M V_i = V_i for each of the these three linear
> subspaces V_i.
>
> The first condition, M V_1 = V_1, is satisfied iff the one basis element
> e_1 of V_1 is carried to a multiple of itself, that is, in the
> expansion M e_1 = Sum m_i1 e_i we must have m_1i = 0 for all i>1.
> (We also need m_11 <> 0 , but that will then be forced by the invertibility
> of M.)
>
> The next condition, M V_2 = V_2, is satisfied iff each of the three
> basis elements e_1, e_2, and e_3 of V_2 is carried back into V_2
> (Again this really only gives M V_2 to be a _subset_ of V_2, but
> the invertibility of M disallows any shrinking of dimensions.) Well,
> we already have M e_1 being a multiple of e_1, but we also need
> M e_2 = m_12 e_1 + m_22 e_2 + m_32 e_3 for some constants m_i2,
> and likewise M e_3. This gives further coefficients of the matrix
> M which must be zero.
>
> Finally the condition M V_3 = V_3 is similarly expressed as the condition
> that M have only zeros in the first five columns, below the first five
> rows.
>
> Thus M must have a characteristic "staircase" upperdiagonal form, the
> diagonal chunks having dimensions 1, 3-1=2, 5-3=2, and 8-5=3. You run into
> this form all the time in this kind of discussion. It shows up a lot in
> texts on homogeneous spaces, Lie Group actions, projective geometry, and
> so on.
>
> Let me know if this is still unclear.
Yes, it is clear to me.
Can the computation of the dimension of the "staircase" upperdiagonal form space
be carried out: 1*1 + 2*3 + 2*5 + 3*8 = 41? After all, it is a simple matter
of counting the non-zero elements in an 8x8 matrix.
Thank you,
Michael Press