dimension of a subspace of a flag variety
$begingroup$
Let $X$ be a topological space. If $X = bigcup U_alpha$ is an open covering of $X$ then $$dim X = sup_alpha dim U_alpha.$$
Now suppose that $X = coprod U_alpha$, i.e., $X$ is the disjoint union of the $U_alpha$'s, but the $U_alpha$'s are NOT open (the most we can assume is that they are locally closed). Can we infer anything about the dimension of $X$ from the dimensions of the disjoint components, other than the dimension of $X$ has to be greater or equal to the dimension of each component?
We can assume that the decomposition is finite, in fact, assume that there just two components.
By dimension I mean the supremum of lengths of chains of closed irreducible subsets, i.e., the notion of dimension used in algebraic geometry for example.
In fact I am interested in a spacial case of the situation above. For me $X = G/B$ is a flag variety. Then $G/B$ has a Schubert cell decomposition. Now suppose that I consider a subspace $Y$ composed of several Schubert cells. I conjecture that the dimension of $Y$ is equal to the dimension of highest-dimensional Schubert cell in $Y$. Is this true?
general-topology algebraic-geometry schubert-calculus
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space. If $X = bigcup U_alpha$ is an open covering of $X$ then $$dim X = sup_alpha dim U_alpha.$$
Now suppose that $X = coprod U_alpha$, i.e., $X$ is the disjoint union of the $U_alpha$'s, but the $U_alpha$'s are NOT open (the most we can assume is that they are locally closed). Can we infer anything about the dimension of $X$ from the dimensions of the disjoint components, other than the dimension of $X$ has to be greater or equal to the dimension of each component?
We can assume that the decomposition is finite, in fact, assume that there just two components.
By dimension I mean the supremum of lengths of chains of closed irreducible subsets, i.e., the notion of dimension used in algebraic geometry for example.
In fact I am interested in a spacial case of the situation above. For me $X = G/B$ is a flag variety. Then $G/B$ has a Schubert cell decomposition. Now suppose that I consider a subspace $Y$ composed of several Schubert cells. I conjecture that the dimension of $Y$ is equal to the dimension of highest-dimensional Schubert cell in $Y$. Is this true?
general-topology algebraic-geometry schubert-calculus
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
$endgroup$
1
$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12
add a comment |
$begingroup$
Let $X$ be a topological space. If $X = bigcup U_alpha$ is an open covering of $X$ then $$dim X = sup_alpha dim U_alpha.$$
Now suppose that $X = coprod U_alpha$, i.e., $X$ is the disjoint union of the $U_alpha$'s, but the $U_alpha$'s are NOT open (the most we can assume is that they are locally closed). Can we infer anything about the dimension of $X$ from the dimensions of the disjoint components, other than the dimension of $X$ has to be greater or equal to the dimension of each component?
We can assume that the decomposition is finite, in fact, assume that there just two components.
By dimension I mean the supremum of lengths of chains of closed irreducible subsets, i.e., the notion of dimension used in algebraic geometry for example.
In fact I am interested in a spacial case of the situation above. For me $X = G/B$ is a flag variety. Then $G/B$ has a Schubert cell decomposition. Now suppose that I consider a subspace $Y$ composed of several Schubert cells. I conjecture that the dimension of $Y$ is equal to the dimension of highest-dimensional Schubert cell in $Y$. Is this true?
general-topology algebraic-geometry schubert-calculus
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
$endgroup$
Let $X$ be a topological space. If $X = bigcup U_alpha$ is an open covering of $X$ then $$dim X = sup_alpha dim U_alpha.$$
Now suppose that $X = coprod U_alpha$, i.e., $X$ is the disjoint union of the $U_alpha$'s, but the $U_alpha$'s are NOT open (the most we can assume is that they are locally closed). Can we infer anything about the dimension of $X$ from the dimensions of the disjoint components, other than the dimension of $X$ has to be greater or equal to the dimension of each component?
We can assume that the decomposition is finite, in fact, assume that there just two components.
By dimension I mean the supremum of lengths of chains of closed irreducible subsets, i.e., the notion of dimension used in algebraic geometry for example.
In fact I am interested in a spacial case of the situation above. For me $X = G/B$ is a flag variety. Then $G/B$ has a Schubert cell decomposition. Now suppose that I consider a subspace $Y$ composed of several Schubert cells. I conjecture that the dimension of $Y$ is equal to the dimension of highest-dimensional Schubert cell in $Y$. Is this true?
general-topology algebraic-geometry schubert-calculus
general-topology algebraic-geometry schubert-calculus
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
edited Jan 5 at 3:16
Matt Samuel
38.1k63767
38.1k63767
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
asked Mar 7 '14 at 21:06
Balerion_the_blackBalerion_the_black
865817
865817
1
$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12
add a comment |
1
$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12
1
1
$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12
$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12
add a comment |
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$begingroup$
By $textrm{dim}$, do you mean the covering dimension?
$endgroup$
– Unwisdom
Mar 7 '14 at 21:12