Talk:Kervaire-Milnor Braid (Ex)

Revision as of 14:59, 4 September 2013

For starters the braid looks as follows:

Filling in the $<wikitex>; For starters the braid looks as follows: $$ \def\curv{1.5pc} \xymatrix{ \pi_8(O) \ar[dr] \ar@/u\curv/^J[rr] && \Omega_8^{fr} \ar[dr] \ar@/u\curv/[rr] && L_8(\Z) \ar[dr]\ar@/u\curv/[rr] &&\Theta_7 \ar[dr]\ar@/u\curv/[rr]&& \pi_6(O) \ar[dr] \ar@/u\curv/^J[rr] && \Omega_6^{fr} \ar[dr] \ar@/u\curv/[rr] && L_6(\Z) \ar[dr]\ar@/u\curv/[rr] &&\Theta_5 \ar[dr]\ar@/u\curv/[rr]&& \pi_{4}(O) \ & \Theta_{8}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_8 \ar[dr] \ar^s[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_7 \ar_s[dr] \ar[ur] && \Theta_{6}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_6 \ar[dr] \ar^s[ur] && \Theta_{5}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_5 \ar_s[dr] \ar[ur] & \ L_9(\Z) \ar[ur] \ar@/d\curv/[rr] && \Theta_8 \ar[ur] \ar@/d\curv/[rr] && \pi_7(O) \ar[ur] \ar@/d\curv/_J[rr] &&\Omega_7^{fr} \ar[ur]\ar@/d\curv/[rr] && L_7(\Z) \ar[ur] \ar@/d\curv/[rr] && \Theta_6 \ar[ur] \ar@/d\curv/[rr] && \pi_5(O) \ar[ur] \ar@/d\curv/_J[rr] &&\Omega_5^{fr} \ar[ur]\ar@/d\curv/[rr] && L_5(\Z) } $$ Filling in the $L$-groups (which were computed by Kervaire and Milnor), the homotopy groups of the orthogonal group (which are given by Bott periodicity), the framed bordism groups (which were shown to be the stable stems by Pontryagin and in low dimensions computed by Serre), and the $J$-homomorphism (which was computed by Adams and Quillen, but can be computed by hand in low dimensions) we arrive at $$ \def\curv{1.5pc} \xymatrix{ \Z/2 \ar[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]\ar@/u\curv/[rr] && \Z \ar@{^{(}->}[dr]\ar@/u\curv/[rr] &&\Theta_7 \ar[dr]\ar@/u\curv/[rr]&& 0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr] \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]\ar@/u\curv/[rr] &&\Theta_5 \ar[dr]\ar@/u\curv/[rr]&& 0 \ & \Theta_{8}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_8 \ar[dr] \ar^{sign}[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_7 \ar[dr] \ar[ur] && \Theta_{6}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_6 \ar[dr] \ar[ur] && \Theta_{5}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_5 \ar[dr] \ar[ur] & \ 0 \ar[ur] \ar@/d\curv/[rr] && \Theta_8 \ar[ur] \ar@/d\curv/[rr] && \Z \ar[ur] \ar@{->>}@/d\curv/[rr] &&\Z/240 \ar[ur]\ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && \Theta_6 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] &&0 \ar[ur]\ar@/d\curv/[rr] && 0 } $$ where we have used $\kappa$ to denote the map induced by the Kervaire invariant $\Omega_6^{fr} \rightarrow \Z/2$. Using that this is surjective ($S^3 \times S^3$ with its Lie-group framing has Kervaire-invariant L$-groups (which were computed by Kervaire and Milnor), the homotopy groups of the orthogonal group (which are given by Bott periodicity), the framed bordism groups (which were shown to be the stable stems by Pontryagin and in low dimensions computed by Serre), and the $J$-homomorphism (which was computed by Adams and Quillen, but can be computed by hand in low dimensions) we arrive at

Tex syntax error

and the fact that the signature of an almost framed $8$-manifold is divisible by $28$ (and $28$ is actually the signature of ???), we obtain the following maps:

The extension at $\Omega_8^{alm}$ is split since it surjects onto $240 \Z$ (right lower map) which is free. Clearing this and the obvious $0$'s, we find:

$$\displaystyle \Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$

which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($28$ having a common factor with both $60$ and $120$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):

Let's first prove the algebra fact from above: Projecting onto the first factor gives a map of short exact sequences:

The conclusion is then immediate from the snake lemma. The only nontrivial left to prove now is that there is only a single isomorphism class for a diagram of the form

with the obvious exactness properties derived from the braid above. We will convert such a diagram to the one displayed in the braid above in several steps. First, the maps between the $\Z$ entries are certainly given by multiplication with $\pm 28$ and $\pm 240$, respectively. The isomorphism of braids induced by changing the signs of the upper and/or lower $\Z$ as necessary converts these into the multiplication with the positive numbers. Since the upper and lower maps $\Z \rightarrow \Z/ 28(240)$ are surjective the isomorphism of braids given by dividing by the image of $1 \in \Z$ in $\Z/28(240)$ converts these maps into the projections. Switching the sign on the $\Z$-summand of $\Z \times \Z/4$ and using the fact from above, we can change the upper map $\Z \rightarrow \Z \times \Z/4$ (which is given by $(\pm 60, ?)$ into the $(60,?)$. The lower map will then be determined to be $(7,?)$ by commutativity of the left square. Our given diagram is thus isomorphic to the following:

By commutativity of the upper triangle, we also know that the only possibilities for the upper map are $(60,1)$ and $(60,3)$ (otherwise the upper map would not be surjective). Multiplying with $-1$ on the $\Z/4$-summand if necessary we can arrange this to be $(60,1)$. Now for the fun part: Composing with the shearing isomorphism $(x,y) \mapsto (x,x+y)$ of $\Z \times \Z/4$ leaves the map $(60,1)$ fixed, since $60$ is divisible by $4$, and changes the map $(7,?)$ to $(7,?+7) = (7,?-1)$. So using the induced isomorphism of braids we can arrange for $(7,?) = (7,0)$ without changing anything fixed before:

The other two maps are now determined: For the lower triangle to commute we must have the lower right map be $(103,?)$ ($103$ being the multiplicative inverse of $7$ modulo $240$. For the composition with the upper left map to be zero, the second component has to be $- 60 \cdot 103 = 60$ modulo $240$. For other sequence to be exact we need the restriction to $\Z/4$ of the unknown map $\Z \times \Z/4 \rightarrow \Z$ injective (the kernel is torsionfree) so it has to be of the form $(x,7)$ or $(x,21)$. For the upper triangle to commute one then finds the following congruences modulo $28$ to be solved:

$$\displaystyle 60x + 7 = 1 \quad \quad \text{ or } \quad \quad 60x + 21 = 1$$

which amounts to finding solutions to

$$\displaystyle 60x + 28w = 7 \quad \quad \text{ or } \quad \quad 60x + 28w = 20$$

which is obviously impossible in the first case. The solutions modulo $28$ of the second equation are given by $5, 12, 19$ and $26$. For the composition with the lower left map to be zero this number has to be divisible by $4$ and only $12$ is. This concludes negotiations.

Tex syntax error

and the fact that the signature of an almost framed $8$-manifold is divisible by $28$ (and $28$ is actually the signature of ???), we obtain the following maps:

The extension at $\Omega_8^{alm}$ is split since it surjects onto $240 \Z$ (right lower map) which is free. Clearing this and the obvious $0$'s, we find:

$$\displaystyle \Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$

which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($28$ having a common factor with both $60$ and $120$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):

Let's first prove the algebra fact from above: Projecting onto the first factor gives a map of short exact sequences:

The conclusion is then immediate from the snake lemma. The only nontrivial left to prove now is that there is only a single isomorphism class for a diagram of the form

with the obvious exactness properties derived from the braid above. We will convert such a diagram to the one displayed in the braid above in several steps. First, the maps between the $\Z$ entries are certainly given by multiplication with $\pm 28$ and $\pm 240$, respectively. The isomorphism of braids induced by changing the signs of the upper and/or lower $\Z$ as necessary converts these into the multiplication with the positive numbers. Since the upper and lower maps $\Z \rightarrow \Z/ 28(240)$ are surjective the isomorphism of braids given by dividing by the image of $1 \in \Z$ in $\Z/28(240)$ converts these maps into the projections. Switching the sign on the $\Z$-summand of $\Z \times \Z/4$ and using the fact from above, we can change the upper map $\Z \rightarrow \Z \times \Z/4$ (which is given by $(\pm 60, ?)$ into the $(60,?)$. The lower map will then be determined to be $(7,?)$ by commutativity of the left square. Our given diagram is thus isomorphic to the following:

By commutativity of the upper triangle, we also know that the only possibilities for the upper map are $(60,1)$ and $(60,3)$ (otherwise the upper map would not be surjective). Multiplying with $-1$ on the $\Z/4$-summand if necessary we can arrange this to be $(60,1)$. Now for the fun part: Composing with the shearing isomorphism $(x,y) \mapsto (x,x+y)$ of $\Z \times \Z/4$ leaves the map $(60,1)$ fixed, since $60$ is divisible by $4$, and changes the map $(7,?)$ to $(7,?+7) = (7,?-1)$. So using the induced isomorphism of braids we can arrange for $(7,?) = (7,0)$ without changing anything fixed before:

The other two maps are now determined: For the lower triangle to commute we must have the lower right map be $(103,?)$ ($103$ being the multiplicative inverse of $7$ modulo $240$. For the composition with the upper left map to be zero, the second component has to be $- 60 \cdot 103 = 60$ modulo $240$. For other sequence to be exact we need the restriction to $\Z/4$ of the unknown map $\Z \times \Z/4 \rightarrow \Z$ injective (the kernel is torsionfree) so it has to be of the form $(x,7)$ or $(x,21)$. For the upper triangle to commute one then finds the following congruences modulo $28$ to be solved:

$$\displaystyle 60x + 7 = 1 \quad \quad \text{ or } \quad \quad 60x + 21 = 1$$

which amounts to finding solutions to

$$\displaystyle 60x + 28w = 7 \quad \quad \text{ or } \quad \quad 60x + 28w = 20$$

which is obviously impossible in the first case. The solutions modulo $28$ of the second equation are given by $5, 12, 19$ and $26$. For the composition with the lower left map to be zero this number has to be divisible by $4$ and only $12$ is. This concludes negotiations.

Tex syntax error

and the fact that the signature of an almost framed $8$-manifold is divisible by $28$ (and $28$ is actually the signature of ???), we obtain the following maps:

The extension at $\Omega_8^{alm}$ is split since it surjects onto $240 \Z$ (right lower map) which is free. Clearing this and the obvious $0$'s, we find:

$$\displaystyle \Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$

which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($28$ having a common factor with both $60$ and $120$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):

Let's first prove the algebra fact from above: Projecting onto the first factor gives a map of short exact sequences:

The conclusion is then immediate from the snake lemma. The only nontrivial left to prove now is that there is only a single isomorphism class for a diagram of the form

with the obvious exactness properties derived from the braid above. We will convert such a diagram to the one displayed in the braid above in several steps. First, the maps between the $\Z$ entries are certainly given by multiplication with $\pm 28$ and $\pm 240$, respectively. The isomorphism of braids induced by changing the signs of the upper and/or lower $\Z$ as necessary converts these into the multiplication with the positive numbers. Since the upper and lower maps $\Z \rightarrow \Z/ 28(240)$ are surjective the isomorphism of braids given by dividing by the image of $1 \in \Z$ in $\Z/28(240)$ converts these maps into the projections. Switching the sign on the $\Z$-summand of $\Z \times \Z/4$ and using the fact from above, we can change the upper map $\Z \rightarrow \Z \times \Z/4$ (which is given by $(\pm 60, ?)$ into the $(60,?)$. The lower map will then be determined to be $(7,?)$ by commutativity of the left square. Our given diagram is thus isomorphic to the following:

By commutativity of the upper triangle, we also know that the only possibilities for the upper map are $(60,1)$ and $(60,3)$ (otherwise the upper map would not be surjective). Multiplying with $-1$ on the $\Z/4$-summand if necessary we can arrange this to be $(60,1)$. Now for the fun part: Composing with the shearing isomorphism $(x,y) \mapsto (x,x+y)$ of $\Z \times \Z/4$ leaves the map $(60,1)$ fixed, since $60$ is divisible by $4$, and changes the map $(7,?)$ to $(7,?+7) = (7,?-1)$. So using the induced isomorphism of braids we can arrange for $(7,?) = (7,0)$ without changing anything fixed before:

The other two maps are now determined: For the lower triangle to commute we must have the lower right map be $(103,?)$ ($103$ being the multiplicative inverse of $7$ modulo $240$. For the composition with the upper left map to be zero, the second component has to be $- 60 \cdot 103 = 60$ modulo $240$. For other sequence to be exact we need the restriction to $\Z/4$ of the unknown map $\Z \times \Z/4 \rightarrow \Z$ injective (the kernel is torsionfree) so it has to be of the form $(x,7)$ or $(x,21)$. For the upper triangle to commute one then finds the following congruences modulo $28$ to be solved:

$$\displaystyle 60x + 7 = 1 \quad \quad \text{ or } \quad \quad 60x + 21 = 1$$

which amounts to finding solutions to

$$\displaystyle 60x + 28w = 7 \quad \quad \text{ or } \quad \quad 60x + 28w = 20$$

which is obviously impossible in the first case. The solutions modulo $28$ of the second equation are given by $5, 12, 19$ and $26$. For the composition with the lower left map to be zero this number has to be divisible by $4$ and only $12$ is. This concludes negotiations.

Tex syntax error

and the fact that the signature of an almost framed $8$-manifold is divisible by $28$ (and $28$ is actually the signature of ???), we obtain the following maps:

The extension at $\Omega_8^{alm}$ is split since it surjects onto $240 \Z$ (right lower map) which is free. Clearing this and the obvious $0$'s, we find:

$$\displaystyle \Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$

which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($28$ having a common factor with both $60$ and $120$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):

Let's first prove the algebra fact from above: Projecting onto the first factor gives a map of short exact sequences:

The conclusion is then immediate from the snake lemma. The only nontrivial left to prove now is that there is only a single isomorphism class for a diagram of the form

with the obvious exactness properties derived from the braid above. We will convert such a diagram to the one displayed in the braid above in several steps. First, the maps between the $\Z$ entries are certainly given by multiplication with $\pm 28$ and $\pm 240$, respectively. The isomorphism of braids induced by changing the signs of the upper and/or lower $\Z$ as necessary converts these into the multiplication with the positive numbers. Since the upper and lower maps $\Z \rightarrow \Z/ 28(240)$ are surjective the isomorphism of braids given by dividing by the image of $1 \in \Z$ in $\Z/28(240)$ converts these maps into the projections. Switching the sign on the $\Z$-summand of $\Z \times \Z/4$ and using the fact from above, we can change the upper map $\Z \rightarrow \Z \times \Z/4$ (which is given by $(\pm 60, ?)$ into the $(60,?)$. The lower map will then be determined to be $(7,?)$ by commutativity of the left square. Our given diagram is thus isomorphic to the following:

By commutativity of the upper triangle, we also know that the only possibilities for the upper map are $(60,1)$ and $(60,3)$ (otherwise the upper map would not be surjective). Multiplying with $-1$ on the $\Z/4$-summand if necessary we can arrange this to be $(60,1)$. Now for the fun part: Composing with the shearing isomorphism $(x,y) \mapsto (x,x+y)$ of $\Z \times \Z/4$ leaves the map $(60,1)$ fixed, since $60$ is divisible by $4$, and changes the map $(7,?)$ to $(7,?+7) = (7,?-1)$. So using the induced isomorphism of braids we can arrange for $(7,?) = (7,0)$ without changing anything fixed before:

The other two maps are now determined: For the lower triangle to commute we must have the lower right map be $(103,?)$ ($103$ being the multiplicative inverse of $7$ modulo $240$. For the composition with the upper left map to be zero, the second component has to be $- 60 \cdot 103 = 60$ modulo $240$. For other sequence to be exact we need the restriction to $\Z/4$ of the unknown map $\Z \times \Z/4 \rightarrow \Z$ injective (the kernel is torsionfree) so it has to be of the form $(x,7)$ or $(x,21)$. For the upper triangle to commute one then finds the following congruences modulo $28$ to be solved:

$$\displaystyle 60x + 7 = 1 \quad \quad \text{ or } \quad \quad 60x + 21 = 1$$

which amounts to finding solutions to

$$\displaystyle 60x + 28w = 7 \quad \quad \text{ or } \quad \quad 60x + 28w = 20$$

which is obviously impossible in the first case. The solutions modulo $28$ of the second equation are given by $5, 12, 19$ and $26$. For the composition with the lower left map to be zero this number has to be divisible by $4$ and only $12$ is. This concludes negotiations.

Tex syntax error

and the fact that the signature of an almost framed $8$-manifold is divisible by $28$ (and $28$ is actually the signature of ???), we obtain the following maps:

The extension at $\Omega_8^{alm}$ is split since it surjects onto $240 \Z$ (right lower map) which is free. Clearing this and the obvious $0$'s, we find:

$$\displaystyle \Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$

which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($28$ having a common factor with both $60$ and $120$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below):

Let's first prove the algebra fact from above: Projecting onto the first factor gives a map of short exact sequences:

The conclusion is then immediate from the snake lemma. The only nontrivial left to prove now is that there is only a single isomorphism class for a diagram of the form

with the obvious exactness properties derived from the braid above. We will convert such a diagram to the one displayed in the braid above in several steps. First, the maps between the $\Z$ entries are certainly given by multiplication with $\pm 28$ and $\pm 240$, respectively. The isomorphism of braids induced by changing the signs of the upper and/or lower $\Z$ as necessary converts these into the multiplication with the positive numbers. Since the upper and lower maps $\Z \rightarrow \Z/ 28(240)$ are surjective the isomorphism of braids given by dividing by the image of $1 \in \Z$ in $\Z/28(240)$ converts these maps into the projections. Switching the sign on the $\Z$-summand of $\Z \times \Z/4$ and using the fact from above, we can change the upper map $\Z \rightarrow \Z \times \Z/4$ (which is given by $(\pm 60, ?)$ into the $(60,?)$. The lower map will then be determined to be $(7,?)$ by commutativity of the left square. Our given diagram is thus isomorphic to the following:

By commutativity of the upper triangle, we also know that the only possibilities for the upper map are $(60,1)$ and $(60,3)$ (otherwise the upper map would not be surjective). Multiplying with $-1$ on the $\Z/4$-summand if necessary we can arrange this to be $(60,1)$. Now for the fun part: Composing with the shearing isomorphism $(x,y) \mapsto (x,x+y)$ of $\Z \times \Z/4$ leaves the map $(60,1)$ fixed, since $60$ is divisible by $4$, and changes the map $(7,?)$ to $(7,?+7) = (7,?-1)$. So using the induced isomorphism of braids we can arrange for $(7,?) = (7,0)$ without changing anything fixed before:

The other two maps are now determined: For the lower triangle to commute we must have the lower right map be $(103,?)$ ($103$ being the multiplicative inverse of $7$ modulo $240$. For the composition with the upper left map to be zero, the second component has to be $- 60 \cdot 103 = 60$ modulo $240$. For other sequence to be exact we need the restriction to $\Z/4$ of the unknown map $\Z \times \Z/4 \rightarrow \Z$ injective (the kernel is torsionfree) so it has to be of the form $(x,7)$ or $(x,21)$. For the upper triangle to commute one then finds the following congruences modulo $28$ to be solved:

$$\displaystyle 60x + 7 = 1 \quad \quad \text{ or } \quad \quad 60x + 21 = 1$$

which amounts to finding solutions to

$$\displaystyle 60x + 28w = 7 \quad \quad \text{ or } \quad \quad 60x + 28w = 20$$

which is obviously impossible in the first case. The solutions modulo $28$ of the second equation are given by $5, 12, 19$ and $26$. For the composition with the lower left map to be zero this number has to be divisible by $4$ and only $12$ is. This concludes negotiations.