An important invariant of a knot in $S^3$ is its *Alexander polynomial*, related also to *Reidemeister torsion*. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?



Alexander polynomial or Reidemeister torsion for knotted surfaces?
Yes for $S^2$, and more generally, depending on what you are after.


Given any torsion module $M$ over the PID $Q[t,t^{-1}]$, the order of $M$ is well defined in 
$Q[t, t^{-1}]$ up to units, in the usual way. Moreover $M\otimes Q(t)=0$. These two facts are at the heart of why the Alexander polynomial is related to Reidemeister torsion. The way it works is that if $C$ is a f.g. chain complex over $Q[t, t^{-1}]$ its homology is torsion iff $C\otimes Q(t)$ is acyclic. Then the two notions of torsion are related by 


$$\prod\_k order(H\_k(C))^{(-1)^k}=\tau(C)$$


If $S^n\subset S^{n+2}$ and $X$ its complement, then $X$ has a $Z$ cover $\tilde{X}$ and hence an exact sequence of chain complexes $0\to C(\tilde{X})\to C(\tilde{X})\to C(X)\to 0.$ The corresponding long exact sequence shows that $H\_k(\tilde{X})$ is torsion for all $k$, since $H\_k(X)=H\_k(S^1)$. 


Setting $\Delta\_k$ to be the order of $H\_k(\tilde{X})$ gives the $k$-th Alexander polynomial. You get Reidemeister torsion equals the multiplicative Euler characteristic: 


$$\tau= \prod \Delta\_k^{(-1)^k}$$


In the case of $S^1\subset S^3$ Poincare duality relates $\Delta\_2$ and $\Delta\_1$, and $\Delta\_0$ is independent of the knot, so you recover Alexander poly "=" Reidemeister torsion. But in general $\tau$ combines all the $\Delta\_k$.


All this (including how to pick the basis of the acyclic complex) is explained in Milnor's article "infinite cyclic covers."


As far as what happens more generally, all this extends, but you have to be careful when the homology is not torsion, which can happen for surfaces in $S^4$. Then you have to view Reidemeister torsion as a function of the homology , etc. Milnor's other Torsion articles are worthwhile reading for these topics. 


