Quadratic forms for surgery

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1 Introduction

Let ${{Stub}} == Introduction == <wikitex>; Let $(f, b) \colon M \to X$ be a [[degree one normal map]] from a manifold of dimension q$. Then the [[surgery kernel]] of $(f, b)$, $K_q(M)$, comes equipped with a subtle and crucial quadratic refinement. This page describes both the algebraic and geometric aspects of such quadratic refinements </wikitex> == Topology == ===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 (f, b) \colon M \to X$ be a degree one normal map from a manifold of dimension $2q$ . Then the surgery kernel of $(f, b)$ , $K_q(M)$ , 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

Let

$\displaystyle (f,b): M \to X$ $\displaystyle (f,b): M \to X$

be a $q$ $q$ -connected normal map, $m =2q$ $m =2q$ , $q$ $q$ odd and $q \ge 3$ $q \ge 3$ . Assume also $X$ $X$ is $1$ $1$ -connected. Let

$\displaystyle I_q(M)$ $\displaystyle I_q(M)$

be the set of regular homotopy classes of immersions $S^q \to M$ $S^q \to M$ which represent elements of the surgery kernel $K_q(M)$ $K_q(M)$ (with respect to the homomorphism $I_q(M) \to K_q(M)$ $I_q(M) \to K_q(M)$ .

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

$\mu' = \mu + {\mathcal O}$ .

1 Definition of the framing obstruction

Each element of $x\in I_q(M)$ is represented by a commutative square

$\displaystyle \SelectTips{cm}{} \xymatrix{ S^q \ar[r]^\phi \ar[d] & M \ar[d]^{f}\\ D^{q+1} \ar[r] & X }$

with $\phi$ an immersion, and a diagram of normal bundle data

$\displaystyle \SelectTips{cm}{} \xymatrix{ S^q \ar[r]^{\nu_\phi} \ar[d] & B\text{O}_q \ar[d]^{f}\\ D^{q+1} \ar[r] & B\text{O} }$

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

Tex syntax error

${\cal O}(x)$ .

2 Definition of the self-intersection obstruction

A generic immersion $\phi\: S^q \to M$ has only double points whose crossings are transversal. $\mu([\phi]) \in {\Bbb Z}_2$ is defined to be the number of double points of $\phi$ taken modulo two. This only depends on $[\phi]=$ regular homotopy class of $\phi$ .

Note that $\mu$ is a quadratic function, i.e.,

$\displaystyle \mu(x+y) = \mu(x) + \mu(y) + x\cdot y$ $\displaystyle \mu(x+y) = \mu(x) + \mu(y) + x\cdot y$

where $x\cdot y$ denotes the intersection pairing applied to $x$ and $y$ considered as elements of $H_q(M;{\Bbb Z}_2)$ .

3 Homotopy Invariance

Theorem 2.1. The function

$\displaystyle \mu': I_q(M) \to \Bbb Z_2$ $\displaystyle \mu': I_q(M) \to \Bbb Z_2$

is homotopy invariant. That is, if $a,b: S^q \to M$ $a,b: S^q \to M$ are immersions representing the same element $x \in K_q(M)$ $x \in K_q(M)$ , then $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ ).

Proof: The homomorphism $I_q(M) \to K_q(M)$ $I_q(M) \to K_q(M)$ is onto and two-to-one. The distinct elements over a given $x \in K_q(M)$ $x \in K_q(M)$ are detected by Browder's framing obstruction

Tex syntax error

$\displaystyle {\cal O} \in \pi_{q}(\text{O}/\text{O}_q) = \Bbb Z_2$

(this uses Smale-Hirsch theory).

Let $a$ $a$ and

Tex syntax error

$b$ be immersions representing these elements. Then $a$ $a$ and

Tex syntax error

$b$ are not regularly homotopic. (Note: when $q\ne 3,7$ $q\ne 3,7$ the normal bundles of $a$ $a$ and

Tex syntax error

$b$ are distinct; when $q=3,7$ $q=3,7$ they are both trivial.) We can assume without loss in generality that

Tex syntax error

${\cal O}(a) = 0$ (so

Tex syntax error

${\cal O}(b) =1$ ). Then $a$ $a$ is a framed immersion.

Case 1: $\mu(a) = 0$ .

If $\mu(a) = 0$ $\mu(a) = 0$ then a result of Whitney (the "Whitney trick") shows that $a$ $a$ is regularly homotopic to a (framed) embedding, so assume that $a$ $a$ is a framed embedding. Whitney's method of introducing a single double point to $a$ $a$ in a coordinate chart yields a new immersion $b'$ $b'$ such that $b'$ $b'$ has one double point and $b'$ $b'$ still represents $x$ $x$ . Then $\mu(b') = 1$ $\mu(b') = 1$ , so $b'$ $b'$ isn't regularly homotopic to $a$ $a$ . It must therefore be regularly homotopic to

Tex syntax error

$b$ . Hence $\mu(b) = 1$ $\mu(b) = 1$ . It follows that $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ in this case.

Case 2: $\mu(a) = 1$ .

In this case $a$ $a$ is regularly homotopic to an immersion with exactly one double point. By introducing another double point we get a $b''$ $b''$ representing $x$ $x$ such that $\mu(b'') = 0$ $\mu(b'') = 0$ . Then $b''$ $b''$ is not regularly homotopic to $a$ $a$ so it must be regularly homotopic to

Tex syntax error

$b$ . Consequently, $\mu(b) = 0$ $\mu(b) = 0$ . Therefore $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ in this case. $\Box$ $\Box$

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 $E_q(M) \to I_q(M)$ .

Corollary 2.2. The function

Tex syntax error

${\cal O}: E_q(M) \to \Bbb Z_2$ factors through $K_q(M)$ $K_q(M)$ .

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 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}} == Algebra == == References == {{#RefList:}} [[Category:Theory]](f, b) \colon M \to X be a degree one normal map from a manifold of dimension $2q$ $2q$ . Then the surgery kernel of $(f, b)$ $(f, b)$ , $K_q(M)$ $K_q(M)$ , 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

Let

$\displaystyle (f,b): M \to X$ $\displaystyle (f,b): M \to X$

be a $q$ $q$ -connected normal map, $m =2q$ $m =2q$ , $q$ $q$ odd and $q \ge 3$ $q \ge 3$ . Assume also $X$ $X$ is $1$ $1$ -connected. Let

$\displaystyle I_q(M)$ $\displaystyle I_q(M)$

be the set of regular homotopy classes of immersions $S^q \to M$ $S^q \to M$ which represent elements of the surgery kernel $K_q(M)$ $K_q(M)$ (with respect to the homomorphism $I_q(M) \to K_q(M)$ $I_q(M) \to K_q(M)$ .

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

$\mu' = \mu + {\mathcal O}$ .

1 Definition of the framing obstruction

Each element of $x\in I_q(M)$ is represented by a commutative square

$\displaystyle \SelectTips{cm}{} \xymatrix{ S^q \ar[r]^\phi \ar[d] & M \ar[d]^{f}\\ D^{q+1} \ar[r] & X }$

with $\phi$ an immersion, and a diagram of normal bundle data

$\displaystyle \SelectTips{cm}{} \xymatrix{ S^q \ar[r]^{\nu_\phi} \ar[d] & B\text{O}_q \ar[d]^{f}\\ D^{q+1} \ar[r] & B\text{O} }$

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

Tex syntax error

${\cal O}(x)$ .

2 Definition of the self-intersection obstruction

A generic immersion $\phi\: S^q \to M$ has only double points whose crossings are transversal. $\mu([\phi]) \in {\Bbb Z}_2$ is defined to be the number of double points of $\phi$ taken modulo two. This only depends on $[\phi]=$ regular homotopy class of $\phi$ .

Note that $\mu$ is a quadratic function, i.e.,

$\displaystyle \mu(x+y) = \mu(x) + \mu(y) + x\cdot y$ $\displaystyle \mu(x+y) = \mu(x) + \mu(y) + x\cdot y$

where $x\cdot y$ denotes the intersection pairing applied to $x$ and $y$ considered as elements of $H_q(M;{\Bbb Z}_2)$ .

3 Homotopy Invariance

Theorem 2.1. The function

$\displaystyle \mu': I_q(M) \to \Bbb Z_2$ $\displaystyle \mu': I_q(M) \to \Bbb Z_2$

is homotopy invariant. That is, if $a,b: S^q \to M$ $a,b: S^q \to M$ are immersions representing the same element $x \in K_q(M)$ $x \in K_q(M)$ , then $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ ).

Proof: The homomorphism $I_q(M) \to K_q(M)$ $I_q(M) \to K_q(M)$ is onto and two-to-one. The distinct elements over a given $x \in K_q(M)$ $x \in K_q(M)$ are detected by Browder's framing obstruction

Tex syntax error

$\displaystyle {\cal O} \in \pi_{q}(\text{O}/\text{O}_q) = \Bbb Z_2$

(this uses Smale-Hirsch theory).

Let $a$ $a$ and

Tex syntax error

$b$ be immersions representing these elements. Then $a$ $a$ and

Tex syntax error

$b$ are not regularly homotopic. (Note: when $q\ne 3,7$ $q\ne 3,7$ the normal bundles of $a$ $a$ and

Tex syntax error

$b$ are distinct; when $q=3,7$ $q=3,7$ they are both trivial.) We can assume without loss in generality that

Tex syntax error

${\cal O}(a) = 0$ (so

Tex syntax error

${\cal O}(b) =1$ ). Then $a$ $a$ is a framed immersion.

Case 1: $\mu(a) = 0$ .

If $\mu(a) = 0$ $\mu(a) = 0$ then a result of Whitney (the "Whitney trick") shows that $a$ $a$ is regularly homotopic to a (framed) embedding, so assume that $a$ $a$ is a framed embedding. Whitney's method of introducing a single double point to $a$ $a$ in a coordinate chart yields a new immersion $b'$ $b'$ such that $b'$ $b'$ has one double point and $b'$ $b'$ still represents $x$ $x$ . Then $\mu(b') = 1$ $\mu(b') = 1$ , so $b'$ $b'$ isn't regularly homotopic to $a$ $a$ . It must therefore be regularly homotopic to

Tex syntax error

$b$ . Hence $\mu(b) = 1$ $\mu(b) = 1$ . It follows that $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ in this case.

Case 2: $\mu(a) = 1$ .

In this case $a$ $a$ is regularly homotopic to an immersion with exactly one double point. By introducing another double point we get a $b''$ $b''$ representing $x$ $x$ such that $\mu(b'') = 0$ $\mu(b'') = 0$ . Then $b''$ $b''$ is not regularly homotopic to $a$ $a$ so it must be regularly homotopic to

Tex syntax error

$b$ . Consequently, $\mu(b) = 0$ $\mu(b) = 0$ . Therefore $\mu'(a) = \mu'(b)$ $\mu'(a) = \mu'(b)$ in this case. $\Box$ $\Box$

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 $E_q(M) \to I_q(M)$ .

Corollary 2.2. The function

Tex syntax error

${\cal O}: E_q(M) \to \Bbb Z_2$ factors through $K_q(M)$ $K_q(M)$ .

3 Algebra

4 References

Quadratic forms for surgery

Revision as of 04:19, 6 April 2011

Contents

1 Introduction

2 Topology

2.1 The 4k+2 dimensional case

1 Definition of the framing obstruction

2 Definition of the self-intersection obstruction

3 Homotopy Invariance

3 Algebra

4 References

2 Topology

2.1 The 4k+2 dimensional case

1 Definition of the framing obstruction

2 Definition of the self-intersection obstruction

3 Homotopy Invariance

3 Algebra

4 References

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@@ Line 22: / Line 22: @@
 * $\mu' = \mu + {\mathcal O}$.
-Let us briefly recall how ${\cal O}$ is defined. Each element of $x\in I_q(M)$ is represented by a commutative square
+====Definition of the framing obstruction====
+Each element of $x\in I_q(M)$ is represented by a commutative square
 $$
 \SelectTips{cm}{}
@@ Line 42: / Line 43: @@
 defines ${\cal O}(x)$.
+====Definition of the self-intersection obstruction====
+A generic immersion $\phi\: S^q \to M$ has only double points whose crossings are transversal. $\mu([\phi]) \in {\Bbb Z}_2$ is defined to be the number
+of double points of $\phi$ taken modulo two. This only depends on $[\phi]=$ regular homotopy class of $\phi$.
+Note that
+$\mu$ is a ''quadratic function,'' i.e.,
+$$
+\mu(x+y) = \mu(x) + \mu(y) + x\cdot y
+$$
+where $x\cdot y$ denotes the intersection pairing applied to $x$ and $y$ considered as elements of $H_q(M;{\Bbb Z}_2)$.
+====Homotopy Invariance====
 {{beginthm|Theorem}} The function $$ \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}}