Cocompact & discrete lattice
$begingroup$
I don't understand a step in the proof of the following proposition:
Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.
Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.
Why is $V/W$ a trivial vector space?
number-theory algebraic-number-theory vector-lattices
asked Jan 14 at 14:16
CYCCYC
972711
$endgroup$
add a comment |
$begingroup$
I don't understand a step in the proof of the following proposition:
Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.
Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.
Why is $V/W$ a trivial vector space?
number-theory algebraic-number-theory vector-lattices
asked Jan 14 at 14:16
CYCCYC
972711
$endgroup$
add a comment |
$begingroup$
I don't understand a step in the proof of the following proposition:
Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.
Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.
Why is $V/W$ a trivial vector space?
number-theory algebraic-number-theory vector-lattices
asked Jan 14 at 14:16
CYCCYC
972711
$endgroup$
I don't understand a step in the proof of the following proposition:
Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.
Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.
Why is $V/W$ a trivial vector space?
number-theory algebraic-number-theory vector-lattices
number-theory algebraic-number-theory vector-lattices
asked Jan 14 at 14:16
CYCCYC
972711
asked Jan 14 at 14:16
CYCCYC
972711
asked Jan 14 at 14:16
CYCCYC
972711
asked Jan 14 at 14:16
CYCCYC
972711
asked Jan 14 at 14:16
CYCCYC
972711
972711
add a comment |
add a comment |
1 Answer
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$begingroup$
The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
$endgroup$
add a comment |
$begingroup$
The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
$endgroup$
add a comment |
$begingroup$
The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
$endgroup$
The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
answered Jan 14 at 14:31
Lee MosherLee Mosher
49k33685
49k33685
add a comment |
add a comment |
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