Ring structure in the Serre spectral sequence
I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)
$$
begin{array}{ccccccc}
mathbb{Z}a & 0 & mathbb{Z}ax_2 & 0 & mathbb{Z}ax_4 & 0 & ldots\
mathbb{Z}1 & 0 & mathbb{Z}x_2 & 0 & mathbb{Z}x_4 & 0 &ldots
end{array}
$$
What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:
The generators for the $mathbb{Z}$'s
in the upper row are $a$ times the
generators in the lower row, because
the product $E_2^{0,q}times E_2^{s,t}to E_2^{s,t+q}$ is just
multiplication of coefficients.
Can someone explain to me what's going on here?
algebraic-topology
asked Apr 29 '11 at 16:19
dstt
484312
add a comment |
I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)
$$
begin{array}{ccccccc}
mathbb{Z}a & 0 & mathbb{Z}ax_2 & 0 & mathbb{Z}ax_4 & 0 & ldots\
mathbb{Z}1 & 0 & mathbb{Z}x_2 & 0 & mathbb{Z}x_4 & 0 &ldots
end{array}
$$
What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:
The generators for the $mathbb{Z}$'s
in the upper row are $a$ times the
generators in the lower row, because
the product $E_2^{0,q}times E_2^{s,t}to E_2^{s,t+q}$ is just
multiplication of coefficients.
Can someone explain to me what's going on here?
algebraic-topology
asked Apr 29 '11 at 16:19
dstt
484312
3
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22
add a comment |
I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)
$$
begin{array}{ccccccc}
mathbb{Z}a & 0 & mathbb{Z}ax_2 & 0 & mathbb{Z}ax_4 & 0 & ldots\
mathbb{Z}1 & 0 & mathbb{Z}x_2 & 0 & mathbb{Z}x_4 & 0 &ldots
end{array}
$$
What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:
The generators for the $mathbb{Z}$'s
in the upper row are $a$ times the
generators in the lower row, because
the product $E_2^{0,q}times E_2^{s,t}to E_2^{s,t+q}$ is just
multiplication of coefficients.
Can someone explain to me what's going on here?
algebraic-topology
asked Apr 29 '11 at 16:19
dstt
484312
I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)
$$
begin{array}{ccccccc}
mathbb{Z}a & 0 & mathbb{Z}ax_2 & 0 & mathbb{Z}ax_4 & 0 & ldots\
mathbb{Z}1 & 0 & mathbb{Z}x_2 & 0 & mathbb{Z}x_4 & 0 &ldots
end{array}
$$
What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:
The generators for the $mathbb{Z}$'s
in the upper row are $a$ times the
generators in the lower row, because
the product $E_2^{0,q}times E_2^{s,t}to E_2^{s,t+q}$ is just
multiplication of coefficients.
Can someone explain to me what's going on here?
algebraic-topology
algebraic-topology
asked Apr 29 '11 at 16:19
dstt
484312
asked Apr 29 '11 at 16:19
dstt
484312
asked Apr 29 '11 at 16:19
dstt
484312
asked Apr 29 '11 at 16:19
dstt
484312
asked Apr 29 '11 at 16:19
dstt
484312
484312
3
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22
add a comment |
3
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22
3
3
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22
add a comment |
1 Answer
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I'll attempt to add to Dylan Wilson's excellent comment.
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(mathbb{Z},1) to P to K(mathbb{Z},2)$. In general for a fibration $F to E to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(mathbb{Z},1) simeq S^1$, the fiber $K(mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(mathbb{Z},1)) = begin{cases} mathbb{Z} & q = 0,1 \ 0 & text{else.}end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(mathbb{Z},2)) otimes H^q(K(mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(mathbb{Z},2))otimes mathbb{Z}$ is determined, by arguing that $E_3 = E_infty$, we immediately know $E_2^{2,1}$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
add a comment |
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I'll attempt to add to Dylan Wilson's excellent comment.
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(mathbb{Z},1) to P to K(mathbb{Z},2)$. In general for a fibration $F to E to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(mathbb{Z},1) simeq S^1$, the fiber $K(mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(mathbb{Z},1)) = begin{cases} mathbb{Z} & q = 0,1 \ 0 & text{else.}end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(mathbb{Z},2)) otimes H^q(K(mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(mathbb{Z},2))otimes mathbb{Z}$ is determined, by arguing that $E_3 = E_infty$, we immediately know $E_2^{2,1}$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
add a comment |
I'll attempt to add to Dylan Wilson's excellent comment.
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(mathbb{Z},1) to P to K(mathbb{Z},2)$. In general for a fibration $F to E to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(mathbb{Z},1) simeq S^1$, the fiber $K(mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(mathbb{Z},1)) = begin{cases} mathbb{Z} & q = 0,1 \ 0 & text{else.}end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(mathbb{Z},2)) otimes H^q(K(mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(mathbb{Z},2))otimes mathbb{Z}$ is determined, by arguing that $E_3 = E_infty$, we immediately know $E_2^{2,1}$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
add a comment |
I'll attempt to add to Dylan Wilson's excellent comment.
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(mathbb{Z},1) to P to K(mathbb{Z},2)$. In general for a fibration $F to E to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(mathbb{Z},1) simeq S^1$, the fiber $K(mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(mathbb{Z},1)) = begin{cases} mathbb{Z} & q = 0,1 \ 0 & text{else.}end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(mathbb{Z},2)) otimes H^q(K(mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(mathbb{Z},2))otimes mathbb{Z}$ is determined, by arguing that $E_3 = E_infty$, we immediately know $E_2^{2,1}$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
I'll attempt to add to Dylan Wilson's excellent comment.
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(mathbb{Z},1) to P to K(mathbb{Z},2)$. In general for a fibration $F to E to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(mathbb{Z},1) simeq S^1$, the fiber $K(mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(mathbb{Z},1)) = begin{cases} mathbb{Z} & q = 0,1 \ 0 & text{else.}end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(mathbb{Z},2)) otimes H^q(K(mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(mathbb{Z},2))otimes mathbb{Z}$ is determined, by arguing that $E_3 = E_infty$, we immediately know $E_2^{2,1}$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
answered Nov 29 at 17:52
Joshua Mundinger
2,4791026
2,4791026
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
add a comment |
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor..
– Neel
Dec 6 at 15:55
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Perhaps a better way to say it is this: if my coefficients are a direct sum $A oplus A’$, then $H^n(X; A oplus A’) = H^n(X;A) oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;mathbb{Z}) oplus H^n(X;mathbb{Z}) = H^n(X; mathbb{Z}) tensor mathbb{Z}^2$.
– Joshua Mundinger
Dec 6 at 16:01
Okay Thank you.
– Neel
Dec 6 at 16:51
Okay Thank you.
– Neel
Dec 6 at 16:51
add a comment |
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3
Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit.
– Dylan Wilson
Apr 29 '11 at 21:39
For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case...
– Bruno Stonek
Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$
– Appliqué
Apr 17 '16 at 13:22