Mixing
Correspondence of Grassmannian cells
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I have shown that the correspondence $f: X rightarrow mathbb{R} oplus X$ defines an embedding of the Grassmannian $mathrm{Gr}_{k}(mathbb{R}^{n+k})$ into $mathrm{Gr}_{k+1}(mathbb{R} oplus mathbb{R}^{n+k})=mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$. Indeed, this identifies an isomorphism that carries an $r$-cell of $mathrm{Gr}_{n}(mathbb{R}^{m})$ onto the $r$-cell of $mathrm{Gr}_{n+1}(mathbb{R}^{m+1})$ corresponding to the same partition. (This is Problem 6-C in Milnor- Stasheff).
From this, I wish to prove that $f$ takes the $left | underline{a} right |$-cell of $mathrm{Gr}_{k}(mathbb{R}^{n+k})$, which corresponds to the given Schubert symbol $underline{a}$, onto the $left | underline{a} right |$-cell of $mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$, which corresponds to the same Schubert symbol $underline{a}$. How might this be done? Any help would be appreciated.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
$endgroup$
add a comment |
$begingroup$
I have shown that the correspondence $f: X rightarrow mathbb{R} oplus X$ defines an embedding of the Grassmannian $mathrm{Gr}_{k}(mathbb{R}^{n+k})$ into $mathrm{Gr}_{k+1}(mathbb{R} oplus mathbb{R}^{n+k})=mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$. Indeed, this identifies an isomorphism that carries an $r$-cell of $mathrm{Gr}_{n}(mathbb{R}^{m})$ onto the $r$-cell of $mathrm{Gr}_{n+1}(mathbb{R}^{m+1})$ corresponding to the same partition. (This is Problem 6-C in Milnor- Stasheff).
From this, I wish to prove that $f$ takes the $left | underline{a} right |$-cell of $mathrm{Gr}_{k}(mathbb{R}^{n+k})$, which corresponds to the given Schubert symbol $underline{a}$, onto the $left | underline{a} right |$-cell of $mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$, which corresponds to the same Schubert symbol $underline{a}$. How might this be done? Any help would be appreciated.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
$endgroup$
add a comment |
$begingroup$
I have shown that the correspondence $f: X rightarrow mathbb{R} oplus X$ defines an embedding of the Grassmannian $mathrm{Gr}_{k}(mathbb{R}^{n+k})$ into $mathrm{Gr}_{k+1}(mathbb{R} oplus mathbb{R}^{n+k})=mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$. Indeed, this identifies an isomorphism that carries an $r$-cell of $mathrm{Gr}_{n}(mathbb{R}^{m})$ onto the $r$-cell of $mathrm{Gr}_{n+1}(mathbb{R}^{m+1})$ corresponding to the same partition. (This is Problem 6-C in Milnor- Stasheff).
From this, I wish to prove that $f$ takes the $left | underline{a} right |$-cell of $mathrm{Gr}_{k}(mathbb{R}^{n+k})$, which corresponds to the given Schubert symbol $underline{a}$, onto the $left | underline{a} right |$-cell of $mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$, which corresponds to the same Schubert symbol $underline{a}$. How might this be done? Any help would be appreciated.
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
$endgroup$
I have shown that the correspondence $f: X rightarrow mathbb{R} oplus X$ defines an embedding of the Grassmannian $mathrm{Gr}_{k}(mathbb{R}^{n+k})$ into $mathrm{Gr}_{k+1}(mathbb{R} oplus mathbb{R}^{n+k})=mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$. Indeed, this identifies an isomorphism that carries an $r$-cell of $mathrm{Gr}_{n}(mathbb{R}^{m})$ onto the $r$-cell of $mathrm{Gr}_{n+1}(mathbb{R}^{m+1})$ corresponding to the same partition. (This is Problem 6-C in Milnor- Stasheff).
From this, I wish to prove that $f$ takes the $left | underline{a} right |$-cell of $mathrm{Gr}_{k}(mathbb{R}^{n+k})$, which corresponds to the given Schubert symbol $underline{a}$, onto the $left | underline{a} right |$-cell of $mathrm{Gr}_{k+1}(mathbb{R}^{n+k+1})$, which corresponds to the same Schubert symbol $underline{a}$. How might this be done? Any help would be appreciated.
algebraic-geometry algebraic-topology schubert-calculus
algebraic-geometry algebraic-topology schubert-calculus
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
edited Jan 5 at 11:23
Matt Samuel
37.8k63665
37.8k63665
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
asked Mar 18 '14 at 23:29
user 3462user 3462
688312
688312
add a comment |
add a comment |
1 Answer
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$begingroup$
First, for clarity, write $f(X) = mathbf{R}oplus X$.
To talk of Schubert cells you must first fix an ordered basis (or a full flag, equivalently). Next you must think of a compatible full flags in $mathbf{R}^n$ and $mathbf{R}^{n+1}$, (this is implicit in your definition of $f$) then the correspondence in Schubert cells will follow.
(Added for clarification)
If $v_1,v_2,ldots,v_{n+k}$ is the ordered basis of $mathbf{R}^{n+k}$, take $v,v_1,v_2,ldots,v_{n+k}$ in that order as basis of $mathbf{R}^{n+k+1}$. So an element of Gr$_k(mathbf{R}^{n+k})$ along with $v$ generates a $k+1$-dim'l subspace, and that is your embedding. The tuple $a=(a_1,a_2,ldots, a_k)$ specifies where the dimension jumps occur for a k-dim'l space intersected with terms of full flag from the ordered basis (giving a Schubert cell). For a non-decreasing seq. of numbers with jumps at specified slots adding a constant to each term means jumps still occur at the same slots.
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
$endgroup$
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First, for clarity, write $f(X) = mathbf{R}oplus X$.
To talk of Schubert cells you must first fix an ordered basis (or a full flag, equivalently). Next you must think of a compatible full flags in $mathbf{R}^n$ and $mathbf{R}^{n+1}$, (this is implicit in your definition of $f$) then the correspondence in Schubert cells will follow.
(Added for clarification)
If $v_1,v_2,ldots,v_{n+k}$ is the ordered basis of $mathbf{R}^{n+k}$, take $v,v_1,v_2,ldots,v_{n+k}$ in that order as basis of $mathbf{R}^{n+k+1}$. So an element of Gr$_k(mathbf{R}^{n+k})$ along with $v$ generates a $k+1$-dim'l subspace, and that is your embedding. The tuple $a=(a_1,a_2,ldots, a_k)$ specifies where the dimension jumps occur for a k-dim'l space intersected with terms of full flag from the ordered basis (giving a Schubert cell). For a non-decreasing seq. of numbers with jumps at specified slots adding a constant to each term means jumps still occur at the same slots.
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
$endgroup$
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
add a comment |
$begingroup$
First, for clarity, write $f(X) = mathbf{R}oplus X$.
To talk of Schubert cells you must first fix an ordered basis (or a full flag, equivalently). Next you must think of a compatible full flags in $mathbf{R}^n$ and $mathbf{R}^{n+1}$, (this is implicit in your definition of $f$) then the correspondence in Schubert cells will follow.
(Added for clarification)
If $v_1,v_2,ldots,v_{n+k}$ is the ordered basis of $mathbf{R}^{n+k}$, take $v,v_1,v_2,ldots,v_{n+k}$ in that order as basis of $mathbf{R}^{n+k+1}$. So an element of Gr$_k(mathbf{R}^{n+k})$ along with $v$ generates a $k+1$-dim'l subspace, and that is your embedding. The tuple $a=(a_1,a_2,ldots, a_k)$ specifies where the dimension jumps occur for a k-dim'l space intersected with terms of full flag from the ordered basis (giving a Schubert cell). For a non-decreasing seq. of numbers with jumps at specified slots adding a constant to each term means jumps still occur at the same slots.
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
$endgroup$
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
add a comment |
$begingroup$
First, for clarity, write $f(X) = mathbf{R}oplus X$.
To talk of Schubert cells you must first fix an ordered basis (or a full flag, equivalently). Next you must think of a compatible full flags in $mathbf{R}^n$ and $mathbf{R}^{n+1}$, (this is implicit in your definition of $f$) then the correspondence in Schubert cells will follow.
(Added for clarification)
If $v_1,v_2,ldots,v_{n+k}$ is the ordered basis of $mathbf{R}^{n+k}$, take $v,v_1,v_2,ldots,v_{n+k}$ in that order as basis of $mathbf{R}^{n+k+1}$. So an element of Gr$_k(mathbf{R}^{n+k})$ along with $v$ generates a $k+1$-dim'l subspace, and that is your embedding. The tuple $a=(a_1,a_2,ldots, a_k)$ specifies where the dimension jumps occur for a k-dim'l space intersected with terms of full flag from the ordered basis (giving a Schubert cell). For a non-decreasing seq. of numbers with jumps at specified slots adding a constant to each term means jumps still occur at the same slots.
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
$endgroup$
First, for clarity, write $f(X) = mathbf{R}oplus X$.
To talk of Schubert cells you must first fix an ordered basis (or a full flag, equivalently). Next you must think of a compatible full flags in $mathbf{R}^n$ and $mathbf{R}^{n+1}$, (this is implicit in your definition of $f$) then the correspondence in Schubert cells will follow.
(Added for clarification)
If $v_1,v_2,ldots,v_{n+k}$ is the ordered basis of $mathbf{R}^{n+k}$, take $v,v_1,v_2,ldots,v_{n+k}$ in that order as basis of $mathbf{R}^{n+k+1}$. So an element of Gr$_k(mathbf{R}^{n+k})$ along with $v$ generates a $k+1$-dim'l subspace, and that is your embedding. The tuple $a=(a_1,a_2,ldots, a_k)$ specifies where the dimension jumps occur for a k-dim'l space intersected with terms of full flag from the ordered basis (giving a Schubert cell). For a non-decreasing seq. of numbers with jumps at specified slots adding a constant to each term means jumps still occur at the same slots.
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
edited Mar 19 '14 at 4:27
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
answered Mar 19 '14 at 0:28
P VanchinathanP Vanchinathan
14.9k12136
14.9k12136
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
add a comment |
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
$begingroup$
Does the correspondence fall out this easily? Could you be a little more explicit?
$endgroup$
– user 3462
Mar 19 '14 at 2:50
add a comment |
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