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edited Dec 20 2011 at 23:10 |
Edit: Let me give it one more try -- I think I've fixed the example.
The following is the simplest simply connected example I know (note that André's example has a large fundamental group).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,y,z, a_1,a_2,b_1,b_2, c_1,c_2,\ldots\rangle,d)$ with $\deg u = \deg v = \deg w = 3, \deg x = \deg y = \deg z =8, \deg a_i = \deg b_i =\deg c_i= 10$ and the following differentials:
$ du=dv=dw=0, dx=uvw, dy=dz=0, da_1=xu+yv, da_2=xu+zw, db_1=xv+zudb_1=xv+yw, $ $ db_2=xv+zu, dc_1=xw+yu, dc_2=xw+zv $ Keep adding generators to kill off all cohomology in degrees above 11. I believe this algebra finally has all Massey products equal to zero. For degree reasons the only possible nontrivial Massey products can be for two classes of degree 3 and one of degree 6, e.g. something of the form $\langle [u], [vw], [v]\rangle$. As far as I can tell all of them vanish.
On the other hand this algebra is not formal. Indeed, suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism, every closed element of $I$ is exact. Now, since $[y]$ and $[z]$ are independent cohomology classes of degree 8 we must have that $I^{\le 8}=\langle x+k_1y+k_2z\rangle$ for some $k_1,k_2\in\mathbb Q$. Then the class $[u(x + k_1y+k_2z)] = [ux] + k_1[u][y]+k_2[uz]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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edited Dec 20 2011 at 21:50 |
Edit: Let me give it one more try -- I think I've fixed the example.
The following is the simplest simply connected example I know (note that André's example has a large fundamental group).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,y,z, a_1,a_2,b_1,b_2, c_1,c_2,\ldots\rangle,d)$ with $\deg u = \deg v = \deg w = 3, \deg x = \deg y = \deg z =8, \deg a_i = \deg b_i =\deg c_i= 10$ and the following differentials: $ $
du=dv=dw=0, dx=uvw, dy=dz=0, da_1=xu+yv, da_2=xu+zw, db_1=xv+zu, $$ $ $db_2=xv+zu, db_2=xv+zu, dc_1=xw+yu, dc_2=xw+zv $$ Keep adding generators to kill off all cohomology in degrees above 11. I believe this algebra finally has all Massey products equal to zero. For degree reasons the only possible nontrivial Massey products can be for two classes of degree 3 and one of degree 6, e.g. something of the form $\langle [u], [vw], [v]\rangle$. As far as I can tell all of them vanish. On the other hand this algebra is not formal. Indeed, suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism, every closed element of $I$ is exact. Now, since $[y]$ and $[z]$ are independent cohomology classes of degree 8 we must have that $I^{\le 8}=\langle x+k_1y+k_2z\rangle$ for some $k_1,k_2\in\mathbb Q$. Then the class $[u(x + k_1y+k_2z)] = [ux] + k_1[u][y]+k_2[uz]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal. Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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edited Dec 20 2011 at 21:43 |
Edit: I just realized that this example is quite wrong because in this model $d\circ d\ne 0$!The example was originally suggested to Let me by Greg Lupton with the same relations but with deg 3 = deg v = 3, deg w = deg x = 5, deg a = deg b = 7. That one is not formal and $d^2=0$ but it turns out that not all Massey products vanish. We thought that this can be fixed by a simple degree shift but that doesn't work as easily as I thought. I'll leave give it here for now as one more try -- I think the example can be I've fixed but it's definitely wrong as stated. The following is the simplest simply connected exampleI know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann). u,v,w,x,y,z, a_1,a_2,b_1,b_2, c_1,c_2,\ldots\rangle,d)$ with deg $\deg u = \deg v = 4, \deg w = 3, \deg x = 7, \deg a y = \deg b z = 10 8, \deg a_i = \deg b_i =\deg c_i= 10$ and the following differentials: $$du=dv=0, dw=0,dx=uv$du=dv=dw=0, da = ux+vwdx=uvw, db = vx+uw$$ dy=dz=0, da_1=xu+yv, da_2=xu+zw, db_1=xv+zu, $$$$db_2=xv+zu, dc_1=xw+yu, dc_2=xw+zv This I believe this algebra finally has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only possible nontrivial Massey products can happen in degree 11 be for products two classes of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to 3 and one of degree 6, e.g. something of the products form $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], vw], [v]\rangle$ both v]\rangle$. As far as I can tell all of which are easily seen to be zerothem vanish. Lastly, it's easy to show that On the other hand this algebra is not formal. Suppose Indeed, suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism, every closed element of $I$ is exact. For degree reasons Now, since $[y]$ and because $w\notin I$ [z]$ are independent cohomology classes of degree 8 we must have that $I^{\le 7}=\langle x+kw,a,b\rangle$ 8}=\langle x+k_1y+k_2z\rangle$ for some $k\in\mathbb k_1,k_2\in\mathbb Q$. Then the class $[u(x + kw)k_1y+k_2z)] = [ux] + k[u][w]$ k_1[u][y]+k_2[uz]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
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edited Dec 19 2011 at 19:19 |
Edit: I just realized that this example is quite wrong because in this model $d\circ d\ne 0$!
The example was originally suggested to me by Greg Lupton with the same relations but with deg 3 = deg v = 3, deg w = deg x = 5, deg a = deg b = 7. That one is not formal and $d^2=0$ but it turns out that not all Massey products vanish. We thought that this can be fixed by a simple degree shift but that doesn't work as easily as I thought. I'll leave it here for now as I think the example can be fixed but it's definitely wrong as stated.
The following is the simplest simply connected example I know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,a,b,\ldots\rangle,d)$
with deg u = deg v = 4, deg w = deg x = 7, deg a = deg b = 10 and the following differentials:
$$du=dv=0, dw=0,dx=uv, da = ux+vw, db = vx+uw$$ Keep adding generators to kill off all cohomology in degrees above 11.
This algebra has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only nontrivial Massey products can happen in degree 11 for products of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to the products $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], [v]\rangle$ both of which are easily seen to be zero.
Lastly, it's easy to show that this algebra is not formal. Suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V, d),0)$. Let $I=\ker \phi$. Since $\phi$ is a quasi-isomorphism every closed element of $I$ is exact.
For degree reasons and because $w\notin I$ we have $I^{\le 7}=\langle x+kw,a,b\rangle$ for some $k\in\mathbb Q$. Then the class $[u(x + kw)] = [ux] + k[u][w]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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edited Dec 19 2011 at 15:12 |
The following is the simplest simply connected example I know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,a,b,\ldots\rangle,d)$
with deg u = deg v = 4, deg w = deg x = 7, deg a = deg b = 10 and the following differentials:
$$du=dv=0, dw=0,dx=uv, da = ux+vw, db = vx+uw$$ Keep adding generators to kill off all cohomology in degrees above 11.
This algebra has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only nontrivial Massey products can happen in degree 11 for products of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to the products $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], [v]\rangle$ both of which are easily seen to be zero.
Lastly, it's easy to show that this algebra is not formal. If you Suppose we have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(\Lambda V),0)$ then it's a simple observation by Deligne, GriffithsV, Morgan and Sullivan (which was however stated and proved slightly incorrectly in their paper) that after possibly modifying the minimal model by choosing a different set of generators the following holds. Set $N=\ker(\phi)\cap V$d),0)$. Then Let $V=N\oplus (V\cap I=\ker \ker d)$ and any phi$. Since $\phi$ is a quasi-isomorphism every closed element in the ideal of $I=N\cdot\Lambda V$ I$ is exact.
For degree reasons and because $N^{\le w\notin I$ we have $I^{\le 7}=\langle w+kx,a,b\rangle$ x+kw,a,b\rangle$ for some $k\in\mathbb Q$. Then the class $[u(w [u(x + kx)kw)] = [uw] ux] + k[u][x]$ k[u][w]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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edited Dec 18 2011 at 23:09 |
The following is the simplest simply connected example I know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,a,b,\ldots\rangle,d)$
with deg u = deg v = 4, deg w = deg x = 7, deg a = deg b = 10 and the following differentials:
$$du=dv=0, dw=0,dx=uv, da = ux+vw, db = vx+uw$$ Keep adding generators to kill off all cohomology in degrees above 11.
This algebra has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only nontrivial Massey products can happen in degree 11 for products of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to the products $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], [v]\rangle$ both of which are easily seen to be zero.
Lastly, it's easy to show that this algebra is not formal. If you have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(V),0)$ H(\Lambda V),0)$ then it's a simple observation by Deligne, Griffiths, Morgan and Sullivan (which was however stated and proved slightly incorrectly in their paper) that after possibly modifying the minimal model by choosing a different set of generators the following holds. Set $N=\ker(\phi)\cap V$. Then $V=N\oplus (V\cap \ker d)$ and any closed element in the ideal $I=N\cdot\Lambda V$ is exact.
For degree reasons $N^{\le 7}=\langle w+kx,a,b\rangle$ for some $k\in\mathbb Q$. Then the class $[u(w + kx)] = [uw] + k[u][x]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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edited Dec 18 2011 at 22:54 |
The following is the simplest simply connected example I know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,a,b,\ldots\rangle,d)$
with deg u = deg v = 4, deg w = deg x = 7, deg a = deg b = 10 and the following differentials:
$$du=dv=0, dw=0,dx=uv, da = ux+vw, db = vx+uw$$ Keep adding generators to kill off all cohomology in degrees above 11.
This algebra has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only nontrivial Massey products can happen in degree 11 for products of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to the products $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], [v]\rangle$ both of which are easily seen to be zero.
Lastly, it's easy to show that this algebra is not formal. If you have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(V),0)$ then it's a simple observation by Deligne, Griffiths, Morgan and Sullivan (which was however stated and proved slightly incorrectly in their paper) that after possibly modifying the minimal model by choosing a different set of generators the following holds. Set $N=\ker(\phi)\cap V$. Then $V=N\oplus (V\cap \ker d)$ and any closed element in the ideal $I=N\cdot\Lambda V$ is exact.
For degree reasons $N^{\le 7}=\langle w+kx,a,b\rangle$ for some $k\in\mathbb Q$. Then the class $[u(w + kx)] = [uw] + k[u][x]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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answered Dec 18 2011 at 22:21 |
The following is the simplest simply connected example I know. It was suggested to me by Greg Lupton (and fixed by Manuel Amann).
Consider the following dga $(\Lambda V,d)=(\Lambda\langle u,v,w,x,a,b,\ldots\rangle,d)$
with deg u = deg v = 4, deg w = deg x = 7, deg a = deg b = 10 and the following differentials:
$$du=dv=0, dw=0,dx=uv, da = ux+vw, db = vx+uw$$ Keep adding generators to kill off all cohomology in degrees above 11.
This algebra has all Massey products equal to zero. This is pretty easy to check. For degree reasons the only nontrivial Massey products can happen in degree 11 for products of degree 4 cohomology classes $\langle [c_1],[c_2],[c_3]\rangle$. Since we must have $[c_1]\cdot [c_2]=[c_2]\cdot [c_3]=0$, elementary linear algebra reduces the situation to the products $\langle [u], [v], [u]\rangle$ and $\langle [v], [u], [v]\rangle$ both of which are easily seen to be zero.
Lastly, it's easy to show that this algebra is not formal. If you have a quasi-isomorphism $\phi:(\Lambda V,d)\to (H(V),0)$ then it's a simple observation by Deligne, Griffiths, Morgan and Sullivan (which was however stated and proved slightly incorrectly in their paper) that after possibly modifying the minimal model by choosing a different set of generators the following holds. Set $N=\ker(\phi)\cap V$. Then any closed element in the ideal $I=N\cdot\Lambda V$ is exact.
For degree reasons $N^{\le 7}=\langle w+kx,a,b\rangle$ for some $k\in\mathbb Q$. Then the class $[u(w + kx)] = [uw] + k[u][x]$ is an element of $I$ which is closed but not exact which means that $(\Lambda V, d)$ is not formal.
Note that this example is spherically generated meaning that its cohomology algebra is generated by closed elements of $V$ (i.e. by duals of the image of Hurewicz homomorphism). It's easy to see that a formal space has a spherically generated minimal model. This provides a very easy criterion for proving that a space is not formal (if you happen to know the minimal model of the space of course). However, as I said the above example is spherically generated and so it is non-formal for slightly more complicated reasons. I don't know of a non-formal example which is not spherically generated but has trivial Massey products.
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