Quadratic forms for surgery
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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
- .
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 .
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 . 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 .