Talk:Kervaire-Milnor Braid (Ex)

Revision as of 11:23, 4 September 2013 (view source) Fabian Hebestreit (Talk | contribs) ← Older edit		Revision as of 11:40, 4 September 2013 (view source) Fabian Hebestreit (Talk | contribs) Newer edit →
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	\xymatrix{		\xymatrix{
	\Z/2 \ar@{^{(}->}[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]^{\neq 0}\ar@/u\curv/[rr]^0 && \Z \ar[dr]\ar@{->>}@/u\curv/[rr] && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&&		\Z/2 \ar@{^{(}->}[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]^{\neq 0}\ar@/u\curv/[rr]^0 && \Z \ar[dr]\ar@{->>}@/u\curv/[rr] && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&&
−	0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^\cong \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]^0\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \\	+	0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^\cong \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \\
	& \Z/2 \times \Z/2 \ar[dr] \ar[ur]^\cong		& \Z/2 \times \Z/2 \ar[dr] \ar[ur]^\cong
	&& \Z \times \Z/2 \ar[dr]_{\cdot \pm 240} \ar^{\cdot \pm 28}[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur]		&& \Z \times \Z/2 \ar[dr]_{\cdot \pm 240} \ar^{\cdot \pm 28}[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur]
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	\def\curv{1.5pc}		\def\curv{1.5pc}
	\xymatrix{		\xymatrix{
−	\Z/2 \ar@{^{(}->}[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]^{\neq 0}\ar@/u\curv/[rr]^0 && \Z \ar[dr]^{(60, 1)}\ar@{->>}@/u\curv/[rr] && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&&	+	\Z/2 \ar[dr]^{(1,0)} \ar@/u\curv/[rr]^{(1,0)} && \Z/2 \times \Z/2 \ar[dr]^{\binom {0 \ 0}{0 \ 1}} \ar@/u\curv/[rr] && \Z \ar[dr]^{(60, 1)}\ar@/u\curv/[rr]^{pr} && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&&
−	0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^\cong \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]^0\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \\	+	0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^{id} \ar@/u\curv/[rr]^{id} && \Z/2 \ar[dr]\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \\
−	& \Z/2 \times \Z/2 \ar[dr] \ar[ur]^\cong	+	& \Z/2 \times \Z/2 \ar[dr]_{\binom 0 1} \ar[ur]^{id}
	&& \Z \times \Z/2 \ar[dr]_{\cdot 240} \ar^{\cdot 28}[ur] && \Z \times \Z/4 \ar[dr]_{\binom {103}{60}} \ar[ur]^{\binom 5{21}}		&& \Z \times \Z/2 \ar[dr]_{\cdot 240} \ar^{\cdot 28}[ur] && \Z \times \Z/4 \ar[dr]_{\binom {103}{60}} \ar[ur]^{\binom 5{21}}
	&& 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur]		&& 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur]
−	&& \Z/2 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur]^\cong	+	&& \Z/2 \ar[dr] \ar[ur]^{id} && 0 \ar[dr] \ar[ur]
	&& 0 \ar[dr] \ar[ur] & \\		&& 0 \ar[dr] \ar[ur] & \\
−	0 \ar[ur] \ar@/d\curv/[rr] && \Z/2 \ar@{^{(}->}[ur] \ar@/d\curv/[rr] && \Z \ar[ur]_{(7, 0)} \ar@{->>}@/d\curv/[rr]	+	0 \ar[ur] \ar@/d\curv/[rr] && \Z/2 \ar[ur]_{(0,1)} \ar@/d\curv/[rr] && \Z \ar[ur]_{(7, 0)} \ar@/d\curv/[rr]_ {pr}
	&&\Z/240 \ar[ur]\ar@/d\curv/[rr] &&		&&\Z/240 \ar[ur]\ar@/d\curv/[rr] &&
	0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr]		0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr]

---

Revision as of 11:40, 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

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 (

Tex syntax error

$S^3 \times S^3$ with its Lie-group framing has Kervaire-invariant $1$)

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:

This in particular implies the smooth Poincaré conjecture in dimension $5$, which is not covered by the usual collary of the $h$-cobordism theorem. The possibilities for $\Theta_7^{fr}$ are $\Z, \Z \times \Z/2$ and $\Z \times \Z/4$, since it is simultaneously an extension of $\Z$ by $\Z/28$ and $\Z/240$ whose common factors are $1,2$ and $4$. We now claim the following facts from algebra, which we prove after giving the final braid: In an extension of the form the map $\Z \rightarrow \Z \times \Z/a$ is always given by multiplication with $\pm b$ in the first factor. Looking now at the triangle above $\Theta_7^{fr}$, we find (for some integers $x,y,z$)

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):

$) and the fact that the signature of an almost framed $-manifold is divisible by $ (and $ is actually the signature of ???), we obtain the following maps: $$ \def\curv{1.5pc} \xymatrix{ \Z/2 \ar@{^{(}->}[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]\ar@/u\curv/[rr]^0 && \Z \ar[dr]\ar@/u\curv/[rr] &&\Theta_7 \ar@{->>}[dr]\ar@/u\curv/[rr]&& 0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^\cong \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]^0\ar@/u\curv/[rr] &&\Theta_5 \ar[dr]\ar@/u\curv/[rr]&& 0 \ & \Theta_{8}^{fr} \ar[dr] \ar[ur]^\cong && \Omega^{alm}_8 \ar[dr] \ar^{28}[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur] && \Omega^{alm}_7 \ar[dr] \ar[ur] && \Theta_{6}^{fr} \ar[dr]_\cong \ar@{^{(}->}[ur]^0 && \Omega^{alm}_6 \ar[dr] \ar[ur]^\cong && \Theta_{5}^{fr} \ar[dr] \ar[ur]^\cong && \Omega^{alm}_5 \ar[dr] \ar[ur] & \ 0 \ar[ur] \ar@/d\curv/[rr] && \Theta_8 \ar[ur] \ar@/d\curv/[rr] && \Z \ar[ur] \ar@{->>}@/d\curv/[rr] &&\Z/240 \ar[ur]_0\ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && \Theta_6 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] &&0 \ar[ur]\ar@/d\curv/[rr] && 0 } $$ The extension at $\Omega_8^{alm}$ is split since it surjects onto 0 \Z$ (right lower map) which is free. Clearing this and the obvious $L$-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

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 (

Tex syntax error

$S^3 \times S^3$ with its Lie-group framing has Kervaire-invariant $1$)

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:

This in particular implies the smooth Poincaré conjecture in dimension $5$, which is not covered by the usual collary of the $h$-cobordism theorem. The possibilities for $\Theta_7^{fr}$ are $\Z, \Z \times \Z/2$ and $\Z \times \Z/4$, since it is simultaneously an extension of $\Z$ by $\Z/28$ and $\Z/240$ whose common factors are $1,2$ and $4$. We now claim the following facts from algebra, which we prove after giving the final braid: In an extension of the form the map $\Z \rightarrow \Z \times \Z/a$ is always given by multiplication with $\pm b$ in the first factor. Looking now at the triangle above $\Theta_7^{fr}$, we find (for some integers $x,y,z$)

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):

$'s, we find: $$ \def\curv{1.5pc} \xymatrix{ \Z/2 \ar@{^{(}->}[dr] \ar@{^{(}->}@/u\curv/[rr] && \Z/2 \times \Z/2 \ar[dr]^{\neq 0}\ar@/u\curv/[rr]^0 && \Z \ar[dr]\ar@{->>}@/u\curv/[rr] && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&& 0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^\cong \ar@/u\curv/[rr]^\cong && \Z/2 \ar[dr]\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \ & \Z/2 \times \Z/2 \ar[dr] \ar[ur]^\cong && \Z \times \Z/2 \ar[dr]_{\cdot \pm 240} \ar^{\cdot \pm 28}[ur] && \Theta_{7}^{fr} \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur] && \Z/2 \ar[dr] \ar[ur]^\cong && 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur] & \ 0 \ar[ur] \ar@/d\curv/[rr] && \Theta_8 \ar[ur] \ar@/d\curv/[rr] && \Z \ar[ur] \ar@{->>}@/d\curv/[rr] &&\Z/240 \ar[ur]\ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] &&0 \ar[ur]\ar@/d\curv/[rr] && 0 } $$ This in particular implies the smooth Poincaré conjecture in dimension $, which is not covered by the usual collary of the $h$-cobordism theorem. The possibilities for $\Theta_7^{fr}$ are $\Z, \Z \times \Z/2$ and $\Z \times \Z/4$, since it is simultaneously an extension of $\Z$ by $\Z/28$ and $\Z/240$ whose common factors are 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

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 (

Tex syntax error

$S^3 \times S^3$ with its Lie-group framing has Kervaire-invariant $1$)

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:

This in particular implies the smooth Poincaré conjecture in dimension $5$, which is not covered by the usual collary of the $h$-cobordism theorem. The possibilities for $\Theta_7^{fr}$ are $\Z, \Z \times \Z/2$ and $\Z \times \Z/4$, since it is simultaneously an extension of $\Z$ by $\Z/28$ and $\Z/240$ whose common factors are $1,2$ and $4$. We now claim the following facts from algebra, which we prove after giving the final braid: In an extension of the form the map $\Z \rightarrow \Z \times \Z/a$ is always given by multiplication with $\pm b$ in the first factor. Looking now at the triangle above $\Theta_7^{fr}$, we find (for some integers $x,y,z$)

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):

,2$ and $. We now claim the following facts from algebra, which we prove after giving the final braid: In an extension of the form $$\Z \rightarrow \Z \times \Z/a \rightarrow \Z/ab$$ the map $\Z \rightarrow \Z \times \Z/a$ is always given by multiplication with $\pm b$ in the first factor. Looking now at the triangle above $\Theta_7^{fr}$, we find (for some integers $x,y,z$) $$\def\curv{1.5pc} \xymatrix{ \Z \ar[dr]^{(\cdot \pm \frac {240} i, x)}\ar@{->>}@/u\curv/[rr] & & \Z/28 \ & \Z \times \Z/i \ar[ur]_{\binom y {z \cdot \frac {240} i}} & }$$ which is only possible for $i = 4$, since the upper map cannot possibly be surjective in the other cases ($ having a common factor with both $ and 0$). With this as input we claim that there is a unique isomorphism class of braids left and it looks as follows (proof given below): $$ \def\curv{1.5pc} \xymatrix{ \Z/2 \ar[dr]^{(1,0)} \ar@/u\curv/[rr]^{(1,0)} && \Z/2 \times \Z/2 \ar[dr]^{\binom {0 \ 0}{0 \ 1}} \ar@/u\curv/[rr] && \Z \ar[dr]^{(60, 1)}\ar@/u\curv/[rr]^{pr} && \Z/28 \ar@{->>}[dr]\ar@/u\curv/[rr]&& 0 \ar[dr] \ar@/u\curv/[rr] && \Z/2 \ar[dr]^{id} \ar@/u\curv/[rr]^{id} && \Z/2 \ar[dr]\ar@/u\curv/[rr] &&0 \ar[dr]\ar@/u\curv/[rr]&& 0 \ & \Z/2 \times \Z/2 \ar[dr]_{\binom 0 1} \ar[ur]^{id} && \Z \times \Z/2 \ar[dr]_{\cdot 240} \ar^{\cdot 28}[ur] && \Z \times \Z/4 \ar[dr]_{\binom {103}{60}} \ar[ur]^{\binom 5{21}} && 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur] && \Z/2 \ar[dr] \ar[ur]^{id} && 0 \ar[dr] \ar[ur] && 0 \ar[dr] \ar[ur] & \ 0 \ar[ur] \ar@/d\curv/[rr] && \Z/2 \ar[ur]_{(0,1)} \ar@/d\curv/[rr] && \Z \ar[ur]_{(7, 0)} \ar@/d\curv/[rr]_ {pr} &&\Z/240 \ar[ur]\ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] && 0 \ar[ur] \ar@/d\curv/[rr] &&0 \ar[ur]\ar@/d\curv/[rr] && 0 } $$ 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

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 (

Tex syntax error

$S^3 \times S^3$ with its Lie-group framing has Kervaire-invariant $1$)

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:

This in particular implies the smooth Poincaré conjecture in dimension $5$, which is not covered by the usual collary of the $h$-cobordism theorem. The possibilities for $\Theta_7^{fr}$ are $\Z, \Z \times \Z/2$ and $\Z \times \Z/4$, since it is simultaneously an extension of $\Z$ by $\Z/28$ and $\Z/240$ whose common factors are $1,2$ and $4$. We now claim the following facts from algebra, which we prove after giving the final braid: In an extension of the form the map $\Z \rightarrow \Z \times \Z/a$ is always given by multiplication with $\pm b$ in the first factor. Looking now at the triangle above $\Theta_7^{fr}$, we find (for some integers $x,y,z$)

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):