Mixing
Can one hear the shape of a DeRham ring ? Is any ring a DeRham ring?
5
$begingroup$
Sorry for the Clicbait title, but take $A$ a graded $mathbb R$ commutative (in the graded sense) algebra of finite dimension.
Does there exist a smooth manifold $M$ having $A$ as a DeRham cohomology ring : $H^*(M) simeq A$ ?
In the negative case, what would be a natural restriction that I missed on the structure of the DeRham ring ?
I was thinking of taking generators for the algebra and taking product of spheres. But i have problems when those generators verfies relations among them.
differential-geometry ring-theory
asked Jan 18 at 9:46
BleuderkBleuderk
1147
$endgroup$
|
show 3 more comments
5
$begingroup$
Sorry for the Clicbait title, but take $A$ a graded $mathbb R$ commutative (in the graded sense) algebra of finite dimension.
Does there exist a smooth manifold $M$ having $A$ as a DeRham cohomology ring : $H^*(M) simeq A$ ?
In the negative case, what would be a natural restriction that I missed on the structure of the DeRham ring ?
I was thinking of taking generators for the algebra and taking product of spheres. But i have problems when those generators verfies relations among them.
differential-geometry ring-theory
asked Jan 18 at 9:46
BleuderkBleuderk
1147
$endgroup$
1
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04
|
show 3 more comments
5
5
5
2
$begingroup$
Sorry for the Clicbait title, but take $A$ a graded $mathbb R$ commutative (in the graded sense) algebra of finite dimension.
Does there exist a smooth manifold $M$ having $A$ as a DeRham cohomology ring : $H^*(M) simeq A$ ?
In the negative case, what would be a natural restriction that I missed on the structure of the DeRham ring ?
I was thinking of taking generators for the algebra and taking product of spheres. But i have problems when those generators verfies relations among them.
differential-geometry ring-theory
asked Jan 18 at 9:46
BleuderkBleuderk
1147
$endgroup$
Sorry for the Clicbait title, but take $A$ a graded $mathbb R$ commutative (in the graded sense) algebra of finite dimension.
Does there exist a smooth manifold $M$ having $A$ as a DeRham cohomology ring : $H^*(M) simeq A$ ?
In the negative case, what would be a natural restriction that I missed on the structure of the DeRham ring ?
I was thinking of taking generators for the algebra and taking product of spheres. But i have problems when those generators verfies relations among them.
differential-geometry ring-theory
differential-geometry ring-theory
asked Jan 18 at 9:46
BleuderkBleuderk
1147
asked Jan 18 at 9:46
BleuderkBleuderk
1147
edited Jan 18 at 9:55
Bleuderk
asked Jan 18 at 9:46
BleuderkBleuderk
1147
asked Jan 18 at 9:46
BleuderkBleuderk
1147
asked Jan 18 at 9:46
BleuderkBleuderk
1147
1147
1
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04
|
show 3 more comments
1
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04
1
1
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04
|
show 3 more comments
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1
$begingroup$
I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' otimes_{Bbb Z} A$ for some finitely generated $Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space.
$endgroup$
– user98602
Jan 19 at 20:51
$begingroup$
So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem
$endgroup$
– Bleuderk
Jan 21 at 10:08
$begingroup$
.... resolving it by the negative
$endgroup$
– Bleuderk
Jan 21 at 10:25
$begingroup$
0) Real cohomology, not $Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $Bbb Z$-algebra, but we do not try to realize the $Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) cong Bbb R^{# text{components of } X}$. This is a product of copies of $Bbb R$. The algebra $Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.)
$endgroup$
– user98602
Jan 21 at 14:47
$begingroup$
In fact that paper seems to give the answer in the positive for rings concentrated in even degrees.
$endgroup$
– user98602
Jan 21 at 15:04