Quadratic forms for surgery

This page has not been refereed. The information given here might be incomplete or provisional.

1 Introduction

Let $(f, b) \colon M \to X$${{Stub}} == Introduction == <wikitex>; Let $(f, b) \colon M \to X$ be a [[degree one normal map]] from a manifold of dimension q$. Then the [[surgery kernel]] of $(f, b)$, $K_q(M)$, comes equipped with a subtle and crucial quadratic refinement. This page describes both the algebraic and geometric aspects of such quadratic refinements </wikitex> == Topology == ===The 4k+2 dimensional case === <wikitex>; Let $$(f,b): M \to X$$ be a $q$-connected normal map, $m =2q$, $q$ odd and $q \ge 3$. Assume also $X$ is (f, b) \colon M \to X$ be a degree one normal map from a manifold of dimension $2q$$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$$q \ge 3$).

Then we have three invariants:

• $\mu =$$\mu =$ double point obstruction $I_q(M) \to \Bbb Z_2$$I_q(M) \to \Bbb Z_2$,

• ${\mathcal O} =$${\mathcal O} =$ Browder's framing obstruction $I_q(M) \to \Bbb Z_2$$I_q(M) \to \Bbb Z_2$, and

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

1 Definition of the framing obstruction

Each element of $x\in I_q(M)$$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$$\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$$\phi$. The homotopy class of the latter diagram defines an element of $\pi_q(\text{O}/\text{O}_q) \in \Bbb Z_2$$\pi_q(\text{O}/\text{O}_q) \in \Bbb Z_2$. This element defines ${\mathcal O}(x)$${\mathcal O}(x)$.

2 Definition of the self-intersection obstruction

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

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

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

(this uses Smale-Hirsch theory).

Let $a$$a$ and $b$$b$ be immersions representing these elements. Then $a$$a$ and $b$$b$ are not regularly homotopic. (Note: when $q\ne 3,7$$q\ne 3,7$ the normal bundles of $a$$a$ and $b$$b$ are distinct; when $q=3,7$$q=3,7$ they are both trivial.) We can assume without loss in generality that ${\mathcal O}(a) = 0$${\mathcal O}(a) = 0$ (so ${\mathcal O}(b) =1$${\mathcal O}(b) =1$). Then $a$$a$ is a framed immersion.

• Case 1: $\mu(a) = 0$$\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$$a$. It must therefore be regularly homotopic to $b$$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$$\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 $b$$b$. Consequently, $\mu(b) = 0$$\mu(b) = 0$. Therefore $\mu'(a) = \mu'(b)$$\mu'(a) = \mu'(b)$ in this case.$\Box$$\Box$

Let $E_q(M)$$E_q(M)$ denote the isotopy classes of embeddings $S^q \to M$$S^q \to M$ representing elements of $K_q(M)$$K_q(M)$. Then we have a function $E_q(M) \to I_q(M)$$E_q(M) \to I_q(M)$.

Corollary 2.2. The function ${\mathcal O}: E_q(M) \to \Bbb Z_2$${\mathcal O}: E_q(M) \to \Bbb Z_2$ factors through $K_q(M)$$K_q(M)$.

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 * $\mu' = \mu + {\mathcal O}$. ====Definition of the framing obstruction==== 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 ${\mathcal O}(x)$. ====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}} '''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 $${\mathcal O} \in \pi_{q}(\text{O}/\text{O}_q) = \Bbb Z_2$$ (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 ${\mathcal O}(a) = 0$ (so ${\mathcal O}(b) =1$). 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 ${\mathcal O}: E_q(M) \to \Bbb Z_2$ 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$$q \ge 3$).

Then we have three invariants:

• $\mu =$$\mu =$ double point obstruction $I_q(M) \to \Bbb Z_2$$I_q(M) \to \Bbb Z_2$,

• ${\mathcal O} =$${\mathcal O} =$ Browder's framing obstruction $I_q(M) \to \Bbb Z_2$$I_q(M) \to \Bbb Z_2$, and

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

1 Definition of the framing obstruction

Each element of $x\in I_q(M)$$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$$\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$$\phi$. The homotopy class of the latter diagram defines an element of $\pi_q(\text{O}/\text{O}_q) \in \Bbb Z_2$$\pi_q(\text{O}/\text{O}_q) \in \Bbb Z_2$. This element defines ${\mathcal O}(x)$${\mathcal O}(x)$.

2 Definition of the self-intersection obstruction

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

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

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

(this uses Smale-Hirsch theory).

Let $a$$a$ and $b$$b$ be immersions representing these elements. Then $a$$a$ and $b$$b$ are not regularly homotopic. (Note: when $q\ne 3,7$$q\ne 3,7$ the normal bundles of $a$$a$ and $b$$b$ are distinct; when $q=3,7$$q=3,7$ they are both trivial.) We can assume without loss in generality that ${\mathcal O}(a) = 0$${\mathcal O}(a) = 0$ (so ${\mathcal O}(b) =1$${\mathcal O}(b) =1$). Then $a$$a$ is a framed immersion.

• Case 1: $\mu(a) = 0$$\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$$a$. It must therefore be regularly homotopic to $b$$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$$\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 $b$$b$. Consequently, $\mu(b) = 0$$\mu(b) = 0$. Therefore $\mu'(a) = \mu'(b)$$\mu'(a) = \mu'(b)$ in this case.$\Box$$\Box$

Let $E_q(M)$$E_q(M)$ denote the isotopy classes of embeddings $S^q \to M$$S^q \to M$ representing elements of $K_q(M)$$K_q(M)$. Then we have a function $E_q(M) \to I_q(M)$$E_q(M) \to I_q(M)$.

Corollary 2.2. The function ${\mathcal O}: E_q(M) \to \Bbb Z_2$${\mathcal O}: E_q(M) \to \Bbb Z_2$ factors through $K_q(M)$$K_q(M)$.