Can one deform an immersion of a 3-manifold in $\mathbb R^4$ to an embedding in $\mathbb R^6$?

Let $M^3$ be an oriented 3-manifold, and let $f:M^3\looparrowright \mathbb R^4$ be a codimension one immersion. Is it possible to find a small deformation of the composite map $$M^3 \to \mathbb R^4 \to \mathbb R^6$$ which is an embedding?

(I expect the answer to be "no", and so I'm mostly interested in the method of proof.)

• Is this true for $M^2\to\mathbb{R}^3\to\mathbb{R}^5$? – Chris Gerig Dec 14 '15 at 18:59
• What category are you interested in? – Igor Rivin Dec 14 '15 at 19:42
• @ChrisGerig: your case follows from the weak Whitney theorem. en.wikipedia.org/wiki/Whitney_embedding_theorem The OP does not. (And the category doesn't matter too much here; though for the OP it may.) – Willie Wong Dec 14 '15 at 19:45
• smooth category. – André Henriques Dec 15 '15 at 13:59

Quoting Theorem F of this paper by Ulrich Koschorke:

For any self-transverse immersion $j$ of a closed 3-manifold $M$ into $\mathbb{R}^4$ the following integers are equal modulo 2:

• the Euler number of the surface of double points of $j$;
• the number of quadruple points of $j$;
• the number of double points of any self-transverse immersion $M\looparrowright\mathbb{R}^6$ which is regularly homotopic to $M \stackrel{j}{\looparrowright}\mathbb{R}^4\subseteq\mathbb{R}^6$.

Moreover, there exists an oriented 3-manifold immersed into $\mathbb{R}^4$ such that all these numbers are odd.

An explicit geometric construction of the required immersion is given in the final section.

• In case anyone is interested, Theorem F of that paper also gives the analog for $M^2\to\mathbb{R}^3\to\mathbb{R}^4$. There is a nice example (due to Giller) for which Koschorke's theorem doesn't help: The oriented double cover of Boy's surface is an immersion of $S^2$ into $\mathbb{R}^3$ that cannot be "lifted" to an embedding in $\mathbb{R}^4$. – Chris Gerig Dec 15 '15 at 1:34
• The oriented 3-manifold immersed in $\mathbb R^4$ such that all these numbers are odd can be constructed as follows. Consider the 2-sphere eversion whose middle point is the Morin surface. This can be viewed as an immersion $S^2\times [0,1] \to \mathbb R^3\times [0,1]$. Now add two half-3-spheres at the end. This yields an immesred $S^3$ with a single quadruple point. I believe that this is due to Carter. – André Henriques Dec 15 '15 at 13:57

The existence of such an immersion is related to the existence of odd Hopf invariant in $\pi_7(S^4)$. Namely take any such element, for example the homotopy class of the Hopf map. Represent it - by the Pontrjagin construction - as an embedded framed 3-dimensional submanifold in $R^7$. Using the so called Compression theorem (by Rourke Sanderson, it is more or less equivalent to Smale Hirsch immersion theory) you may isotope this framed immersion so, that one of the framing normal vectors becomes everywhere parallel to the 7-th coordinate direction of $R^7$. Then projecting it to $R^6$ we obtain a framed immersion in $R^6$. The (algebraic) number of double points (they have signs, since the double branches have an ordering, which was higher, which was lower) agrees with the Hopf invariant of the original map. Applying two more times the compression, you can push the immersion into $R^4$ as an immersion. When you lift back from $R^4$ to $R^6$ after arbitrary regular homotopy, the pairity of the double points remains unchanged. (This is the stable Hopf invariant of the element of the stable homotopy class in $\pi^s(3)$ represented by the obtained immersion.)

• Very interesting! What is the analog statement for the other Hopf invariant one elements, in $\pi_3(S^2)$ and in $\pi_{15}(S^8)$? – André Henriques Dec 15 '15 at 21:58
• Similarly you can produce an immersion of a 7-manifold into $R^8$, that can not be lifted as an embedding into $R^{14}$. With $\pi_3(S^2)$ you can not play this game, there is no enough dimension for it. One can say, that projecting the Hopf map into an immersion in the plane (after makig one normal vector vertical) you will have algebraically one double point. – András Szűcs Dec 15 '15 at 22:04
• Can you explain what 3-manifold is represented by Hopf map $S^7 \to S^4$ ? Fiber of Hopf map is just 3-sphere, point on quaternion projective space $\mathbb H P^1 = S^4$. – user21230 Mar 3 '16 at 10:15