Kervaire invariant

An invariant of an almost-parallelizable smooth manifold $M$ of dimension $k + 2$, defined as the Arf-invariant of the quadratic form modulo 2 on the lattice of the $( 2 k + 1 )$-dimensional homology space of $M$.

Let $M$ be a simply-connected almost-parallelizable closed smooth manifold of dimension $4 k + 2$ whose homology groups $H _ {i} ( M ; \mathbf Z )$ vanish for $0 < i < 4 k + 2$, except for $V = H _ {2k+ 1} ( M ; \mathbf Z )$.

On the free Abelian group $V$ there is a skew-symmetric intersection form of cycles $\Phi ( x , y )$, $\Phi : V \times V \rightarrow \mathbf Z$, and the dimension of the integral lattice in $V$ is equal to $2 m$. There exists on $V$ a function $\Phi _ {0} : V \rightarrow \mathbf Z _ {2}$ defined as follows: If $x \in V$, then there exists a smooth imbedding of the sphere $S ^ {2k+ 1}$ into $M$ that realizes the given element $x$, $k \geq 1$. A tubular neighbourhood of this sphere $S ^ {2k+ 1}$ in $M$ is parallelizable, and it can be either trivial or isomorphic to a tubular neighbourhood of the diagonal in the product $S ^ {2k+ 1} \times S ^ {2k+ 1}$. Here, the tubular neighbourhood of the diagonal in $S ^ {2k+ 1} \times S ^ {2k+ 1}$ is non-trivial if and only if $2 k + 1 \neq 1 , 3 , 7$ (see Hopf invariant). The value of $\Phi _ {0}$ is zero or one depending on whether or not the tubular neighbourhood of $S ^ {2k+ 1}$ realizing $x$ in $M$ is trivial, $2 k + 1 \neq 1 , 3 , 7$. The function $\Phi _ {0} : V \rightarrow \mathbf Z _ {2}$ satisfies the condition

$$\Phi _ {0} ( x + y ) \equiv \Phi _ {0} ( x) + \Phi _ {0} ( y) + \Phi ( x , y ) \mathop{\rm mod} 2 .$$

The Arf-invariant of $\Phi _ {0}$ is also called the Kervaire invariant of the manifold $M ^ {4k+ 2}$, $2 k + 1 \neq 1 , 3 , 7$.

If the Kervaire invariant of $M ^ {4k+ 2}$ is equal to zero, then there exists a symplectic basis $( e _ {i} , f _ {i} )$ for $V$ such that $\Phi _ {0} ( e _ {i} ) = \Phi _ {0} ( f _ {i} ) = 0$. In this case $M ^ {4k+ 2}$ is a connected sum of a product of spheres

$$M ^ {4k+ 2} = \ ( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {1} \# \dots \# ( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {m} .$$

If, on the other hand, the Kervaire invariant of $M ^ {4k+ 2}$ is non-zero, then there is a symplectic basis $( e _ {i} , f _ {i} )$ for $V$ such that $\Phi _ {0} ( e _ {i} ) = \Phi _ {0} ( f _ {i} ) = 0$ for $i \neq 1$ and $\Phi _ {0} ( e _ {1} ) = \Phi _ {0} ( f _ {1} ) = 1$. In this case the union of the tubular neighbourhoods of the two $( 2 k + 1 )$- dimensional spheres, imbedded in $M ^ {4k+ 2}$ with transversal intersection at a point and realizing the elements $e _ {1}$, $f _ {1}$, gives a manifold $K ^ {4k+ 2}$. It is called the Kervaire manifold (see Dendritic manifold); its boundary $\partial K ^ {4k+ 2}$ is diffeomorphic to the standard sphere, while the manifold $M ^ {4k+ 2}$ itself can be expressed as the connected sum

$$M ^ {4k+ 2} = \ \widehat{K} {} ^ {4k+ 2} \# ( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {1} \# \dots$$

$${} \dots \# ( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {m- 1} ,$$

where the smooth closed manifold $\widehat{K} {} ^ {4k+ 2}$ is obtained from $K ^ {4k+ 2}$ by attaching a cell.

If $M ^ {4k+ 2}$, $k \neq 0 , 1 , 3$, is a smooth parallelizable $( 2 k )$-connected manifold with a boundary that is homotopic to a sphere, then the Kervaire invariant of $M ^ {4k+ 2}$ is defined exactly as above and will have the same properties with the difference that, in the decomposition of $M ^ {4k+ 2}$ into a connected sum of simple manifolds, the component $K _ {0} ^ {4k+ 2}$ that is the Kervaire manifold has boundary $\partial K ^ {4k+ 2} = \partial M ^ {4k+ 2}$ (which generally is not diffeomorphic to the standard sphere).

In the cases $k = 0 , 1 , 3$ the original manifolds $M ^ {2}$, $M ^ {6}$, $M ^ {14}$ can be expressed as the connected sum $( S ^ {2k+ 1} \times S ^ {2k+ 1} ) \# \dots \# ( S ^ {2k+ 1} \times S ^ {2k+ 1} )$( if the boundary is empty) or $( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {0} \# \dots \# ( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {m- 1}$( if the boundary is non-empty), where $( S ^ {2k+ 1} \times S ^ {2k+ 1} ) _ {0}$ is obtained by removing an open cell from $S ^ {2k+ 1} \times S ^ {2k+ 1}$.

However, a Kervaire invariant can be defined for the closed manifolds $M ^ {2}$, $M ^ {6}$, $M ^ {14}$ (see Pontryagin invariant; Kervaire–Milnor invariant) and depends in these dimensions on the choice of the framing, that is, it is an invariant of the framed surgery of the pair $( M ^ {4k+ 2} , f _ {r} )$, $k = 0 , 1 , 3$. In dimensions $k \neq 0 , 1 , 3$ the manifold $M ^ {4k+ 2}$ can be modified to the sphere $S ^ {4k+ 2}$ if and only if the pair $( M ^ {4k+ 2} , f _ {r} )$ has a framed surgery to the pair $( S ^ {4k+ 2} , f _ {r} )$ under any choice of $f _ {r}$ on the original manifold $M ^ {4k+ 2}$ (see Surgery on a manifold).

The Kervaire invariant is defined for any stably-parallelizable manifold $M ^ {4k+ 2}$ as an invariant of framed surgery, and any element in the stable homotopy groups of spheres can be represented either as a framed homotopy sphere or as a closed smooth framed Kervaire manifold (in this case $m = 4 k + 2$, $k \neq 0 , 1 , 3$), or as the framed manifold $S ^ {2k+ 1} \times S ^ {2k+ 1}$ if $k = 0 , 1 , 3$.

In other words, the Kervaire invariant can be regarded as an obstruction to "carrying over" the given framing on the manifold to the sphere of the same dimension, $k \neq 0 , 1 , 3$. In this sense the Kervaire invariant fulfills the same role for the values $k = 0 , 1 , 3$: The given framing on $S ^ {2k+ 1} \times S ^ {2k+ 1}$, $k = 0 , 1 , 3$, cannot, in general, be "carried over" to the sphere $S ^ {4k+ 2}$, $k = 0 , 1 , 3$, by means of framed surgery.

L.S. Pontryagin was the first to construct such a framing on the manifold $S ^ {2k+ 1} \times S ^ {2k+ 1}$ for the case $k = 0$, that is, a framing on the $2$- dimensional torus $( ( S ^ {1} \times S ^ {1} ) , f _ {r} )$ that cannot be "carried over" to $S ^ {2}$. There are also such examples of a framing on the manifolds $S ^ {3} \times S ^ {3}$ and $S ^ {7} \times S ^ {7}$.

The fundamental problem concerning the Kervaire invariant is the following: For which odd values of $n$ does there exist a pair $( M ^ {2n} , f _ {r} )$ with non-zero Kervaire invariant? For $n \neq 2 ^ {i} - 1$ the answer to this question is negative and for $n = 2 ^ {i} - 1$ it is affirmative, where $i = 1$ (Pontryagin, see [2]), $i = 2 , 3$ (M.A. Kervaire and J.W. Milnor, [5], [6]), $i = 4$ (W. Browder, [3]), $i = 5 , 6$ (M. Barratt, M. Mahowald, A. Milgram). For other values of $i$ the answer is unknown (1989).

References

[1]	S.P. Novikov, "Homotopy-equivalent smooth manifolds I" Izv. Akad. Nauk SSSR. Ser. Mat. , 28 : 2 (1964) pp. 365–474 (In Russian)
[2]	L.S. Pontryagin, "Smooth manifolds and their applications in homology theory" , Moscow (1976) (In Russian)
[3]	W. Browder, "The Kervaire invariant of framed manifolds and its generalization" Ann. of Math. , 90 (1969) pp. 157–186
[4]	W.B. Browder, "Surgery on simply-connected manifolds" , Springer (1972)
[5]	M. Kervaire, "A manifold which does not admit any differentiable structure" Comm. Math. Helv. , 34 (1960) pp. 257–270
[6]	M.A. Kervaire, J.W. Milnor, "Groups of homotopy spheres I" Ann. Mat. , 77 : 3 (1963) pp. 504–537