Quadratic forms for surgery
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== Topology == | == Topology == | ||
− | Let $$(f,b) | + | |
+ | ===The 4k+2 dimensional case === | ||
+ | <wikitex>; | ||
+ | |||
+ | |||
+ | Let $$(f,b): M \to X$$ be a $q$-connected normal map, $m =2q$, $q$ odd and $q \ge 3$. Assume also $X$ is $1$-connected. Let $$I_q(M)$$ be the set of regular homotopy classes of immersions $S^q \to M$ which represent elements of the surgery kernel $K_q(M)$ (with respect to the homomorphism $I_q(M) \to K_q(M)$. | ||
+ | It is an abelian group using the connected summ operation (this uses the condition $q \ge 3$). | ||
+ | |||
+ | Then we have three invariants: | ||
+ | |||
+ | |||
+ | * $\mu =$ double point obstruction $I_q(M) \to \Bbb Z_2$, | ||
+ | |||
+ | * ${\mathcal O} =$ Browder's framing obstruction $I_q(M) \to \Bbb Z_2$, and | ||
+ | |||
+ | * $\mu' = \mu + {\mathcal O}$. | ||
+ | |||
+ | Let us briefly recall how ${\cal O}$ is defined. Each element of $x\in I_q(M)$ is represented by a commutative square | ||
+ | $$ | ||
+ | \SelectTips{cm}{} | ||
+ | \xymatrix{ | ||
+ | S^q \ar[r]^\phi \ar[d] & M \ar[d]^{f}\\ | ||
+ | D^{q+1} \ar[r] & X | ||
+ | } | ||
+ | $$ | ||
+ | with $\phi$ an immersion, and a diagram of normal bundle data | ||
+ | $$ | ||
+ | \SelectTips{cm}{} | ||
+ | \xymatrix{ | ||
+ | S^q \ar[r]^{\nu_\phi} \ar[d] & B\text{O}_q \ar[d]^{f}\\ | ||
+ | D^{q+1} \ar[r] & B\text{O} | ||
+ | } | ||
+ | $$ | ||
+ | the latter defining a stable trivialization of the normal bundle of $\phi$. | ||
+ | The homotopy class of the latter diagram defines an element of $\pi_q(\text{O}/\text{O}_q) \in \Bbb Z_2$. This element | ||
+ | defines ${\cal O}(x)$. | ||
+ | |||
+ | {{beginthm|Theorem}} The function $$ \mu': I_q(M) \to \Bbb Z_2 $$ is homotopy invariant. That is, if $a,b: S^q \to M$ are immersions representing the same element $x \in K_q(M)$, then $\mu'(a) = \mu'(b)$). | ||
+ | {{endthm}} | ||
+ | |||
+ | |||
+ | |||
+ | '''Proof:''' The homomorphism $I_q(M) \to K_q(M)$ is onto and two-to-one. The distinct elements over a given $x \in K_q(M)$ are detected by Browder's framing obstruction $${\cal O} \in \pi_{q}(\text{O}/\text{O}_q) = \Bbb Z_2$$ | ||
+ | (this uses Smale-Hirsch theory). | ||
+ | |||
+ | Let $a$ and $b$ be immersions representing these elements. Then $a$ and $b$ are not regularly homotopic. (Note: when $q\ne 3,7$ the normal bundles of $a$ and $b$ are distinct; when $q=3,7$ they are both trivial.) We can assume without loss in generality that ${\cal O}(a) = 0$ (so ${\cal O}(b) =1$). Then $a$ is a framed immersion. | ||
+ | |||
+ | * ''Case 1:'' $\mu(a) = 0$. | ||
+ | |||
+ | If $\mu(a) = 0 $ then a result of Whitney (the "Whitney trick") shows that $a$ is regularly homotopic to a (framed) embedding, so assume that $a$ is a framed embedding. Whitney's method of introducing a single double point to $a$ in a coordinate chart yields a new immersion $b'$ such that $b'$ has one double point and $b'$ still represents $x$. Then $\mu(b') = 1$, so $b'$ isn't regularly homotopic to $a$. It must therefore be regularly homotopic to $b$. Hence $\mu(b) = 1$. It follows that $\mu'(a) = \mu'(b)$ in this case. | ||
+ | |||
+ | |||
+ | *''Case 2:'' $\mu(a) = 1$. | ||
+ | |||
+ | In this case $a$ is regularly homotopic to an immersion with exactly one double point. By introducing another double point we get a $b''$ representing $x$ such that $\mu(b'') = 0$. Then $b''$ is not regularly homotopic to $a$ so it must be regularly homotopic to $b$. Consequently, $\mu(b) = 0$. Therefore $\mu'(a) = \mu'(b)$ in this case.$\Box$ | ||
+ | |||
+ | |||
+ | Let $E_q(M)$ denote the isotopy classes of embeddings $S^q \to M$ representing elements of $K_q(M)$. Then we have a function $E_q(M) \to I_q(M)$. | ||
+ | |||
+ | |||
+ | {{beginthm|Corollary}} The function ${\cal O}: E_q(M) \to \Bbb Z_2$ factors through $K_q(M)$. {{endthm}} | ||
+ | </wikitex> | ||
== Algebra == | == Algebra == |
Revision as of 04:01, 6 April 2011
This page has not been refereed. The information given here might be incomplete or provisional. |
Contents |
1 Introduction
Let be a degree one normal map from a manifold of dimension . Then the surgery kernel of , , comes equipped with a subtle and crucial quadratic refinement. This page describes both the algebraic and geometric aspects of such quadratic refinements
2 Topology
2.1 The 4k+2 dimensional case
It is an abelian group using the connected summ operation (this uses the condition ).
Then we have three invariants:
- double point obstruction ,
- Browder's framing obstruction , and
- .
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with an immersion, and a diagram of normal bundle data
the latter defining a stable trivialization of the normal bundle of . The homotopy class of the latter diagram defines an element of . This element
definesTex syntax error.
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(this uses Smale-Hirsch theory).
Let andTex syntax errorbe immersions representing these elements. Then and
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Tex syntax errorare distinct; when they are both trivial.) We can assume without loss in generality that
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- Case 1: .
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- Case 2: .
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3 Algebra
4 References
$-connected. Let $$I_q(M)$$ be the set of regular homotopy classes of immersions $S^q \to M$ which represent elements of the surgery kernel $K_q(M)$ (with respect to the homomorphism $I_q(M) \to K_q(M)$. It is an abelian group. We have two invariants: \roster \item $\mu =$ double point obstruction $I_q(M) \to \Bbb Z_2$ \item $\Cal O =$ Browder's framing obstruction $I_q(M) \to \Bbb Z_2$ \endroster \proclaim{Claim} The function $$ \mu' := \mu + \Cal O\: I_q(M) \to \Bbb Z_2 $$ is homotopy invariant (i.e., if $a,b\: S^q \to M$ are immersions representing the same element $x \in K_q(M)$, then $\mu'(a) = \mu'(b)$). \endproclaim \demo{Proof} The homomorphism $I_q(M) \to K_q(M)$ is onto and two-to-one. The distinct elements over a given $x \in K_q(M)$ are detected by Browder's framing obstruction $\Cal O \in \pi_{q}(\text{O}/\text{O}_q) = \Bbb Z_2$ (this uses Smale-Hirsch theory). Let $a$ and $b$ be these immersions. Then $a$ and $b$ are not regularly homotopic. (Note: when $q\ne 3,7$ the normal bundles of $a$ and $b$ are distinct; when $q=3,7$ they are both trivial.) We can assume without loss in generality that $\Cal O(a) = 0$ (so $\Cal O(b) =1$). Then $a$ is a framed immersion. \medskip \noindent {\sl Case 1:} $\mu(a) = 0$. \medskip If $\mu(a) = 0 $ then a result of Whitney (the ``Whitney trick'') shows that $a$ is regularly homotopic to a (framed) embedding, so assume that $a$ is a framed embedding. Whitney's method of introducing a single double point to $a$ in a coordinate chart yields a new immersion $b'$ such that $b'$ has one double point and $b'$ still represents $x$. Then $\mu(b') = 1$, so $b'$ isn't regularly homotopic to $a$. It must therefore be regularly homotopic to $b$. Hence $\mu(b) = 1$. It follows that $\mu'(a) = \mu'(b)$ in this case. \medskip \noindent {\sl Case 2:} $\mu(a) = 1$. \medskip In this case $a$ is regularly homotopic to an immersion with exactly one double point. By introducing another double point we get a $b''$ representing $x$ such that $\mu(b'') = 0$. Then $b''$ is not regularly homotopic to $a$ so it must be regularly homotopic to $b$. Consequently, $\mu(b) = 0$. Therefore $\mu'(a) = \mu'(b)$ in this case.\qed \enddemo Let $$E_q(M) $$ be the isotopy classes of embeddings $S^q \to M$ representing elements of $K_q(M)$. Then we have a function $E_q(M) \to I_q(M)$. \proclaim{Corollary} The function $\Cal O\: E_q(M) \to \Bbb Z_2$ is homotopy invariant, i.e., it factors through $K_q(M)$. \endproclaim == Algebra == == References == {{#RefList:}} [[Category:Theory]](f, b) \colon M \to X be a degree one normal map from a manifold of dimension . Then the surgery kernel of , , comes equipped with a subtle and crucial quadratic refinement. This page describes both the algebraic and geometric aspects of such quadratic refinements2 Topology
2.1 The 4k+2 dimensional case
It is an abelian group using the connected summ operation (this uses the condition ).
Then we have three invariants:
- double point obstruction ,
- Browder's framing obstruction , and
- .
Tex syntax erroris defined. Each element of is represented by a commutative square
with an immersion, and a diagram of normal bundle data
the latter defining a stable trivialization of the normal bundle of . The homotopy class of the latter diagram defines an element of . This element
definesTex syntax error.
Tex syntax error
(this uses Smale-Hirsch theory).
Let andTex syntax errorbe immersions representing these elements. Then and
Tex syntax errorare not regularly homotopic. (Note: when the normal bundles of and
Tex syntax errorare distinct; when they are both trivial.) We can assume without loss in generality that
Tex syntax error(so
Tex syntax error). Then is a framed immersion.
- Case 1: .
Tex syntax error. Hence . It follows that in this case.
- Case 2: .
Tex syntax error. Consequently, . Therefore in this case.
Let denote the isotopy classes of embeddings representing elements of . Then we have a function .
Tex syntax errorfactors through .