# Can a $W^{1,2}$ map from the disk to the circle restrict to a degree one map on the boundary?

The restriction of a continuous map $D^2\to S^1$ to $\partial D^2\to S^1$ must have degree zero. Is that statement true or false if the map is only $W^{1,2}(D^2;S^1)$ and continuous on $\partial D^2$?

In two dimensions, the Sobolev space $W^{1,2}$ is at the borderline regularity and does not embed into the space of continuous functions. However, we can ask that a map's values be on the circle almost everywhere, giving us the space $W^{1,2}(D^2;S^1)$. Such a map has a well-defined restriction to the boundary; the restriction is a $W^{1/2,2}(\partial D^2;S^1)$ map. If it so happens that this restriction is continuous, we can ask about its degree.

A standard example of a discontinuous $W^{1,2}$ function is $f=\log(\log(4/r))$, and we can use it to construct a discontinuous $W^{1,2}$ circle-valued map $g=e^{if}$. However, this map has degree zero on the boundary, and it's not clear to me if one can do a different construction to get a degree one map.

Another good example to consider is $e^{i\theta}$, which is a $W^{1,p}(D^2;S^1)$ map for any $p<2$, and has degree one on the boundary.

• By the way, continuous or not, $H^{1/2}$ functions on the unit circle always have a well defined degree, as I learned recently in the context of this MO question: mathoverflow.net/questions/255343/… – Christian Remling Feb 14 '17 at 2:08
• @ChristianRemling In fact the question has a positive answer for maps in $W^{1,n}(B^n,S^{n-1})$, see my answer below. – Piotr Hajlasz May 6 '18 at 19:00

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \operatorname{dim}M$, due to Schoen and Uhlenbeck [3] (see also Theorem 2.1 in [1]).

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. Let $B^n=B^n(0,1)$ and let $B^n_{1+\delta}=B^n(0,1+\delta)$ for some small $\delta>0$. First of all, we can extend the mapping $f$ to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$ if $\delta>0$ is small enough. Indeed, the Sobolev extension operator $E$ is defined through averages, see [2], so extending $f$ to $B^n_{1+\delta}\setminus B^n$ gives a function $Ef\in W^{1,n}(B^n_{1+\delta},\mathbb{R}^n)$ that is continuous in the annulus $B^n_{1+\delta}\setminus B^n$. If $\delta>0$ is small enough, $|Ef|>1/2$ on $B^n_{1+\delta}\setminus B^n$ (by continuity and the fact that $|f|=1$ on $\partial B^n$) and hence $\tilde{f}=Ef/|Ef|$ in $B^n_{1+\delta}\setminus B^n$ and $\tilde{f}=f$ in $B^n$ belongs to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$. If we prove that the degree of $\tilde{f}$ on the boundary of $B^n_{1+\delta}$ is zero, then also degree of $f$ on the boundary of $B^n$ is zero (by homotopy invariace of degree and continuity of $\tilde{f}$ in $B^n_{1+\delta}\setminus B^n$).

The above construction shows that we can assume that $f$ is continuous in a neighborhood of $\partial B^n$ (because $\tilde{f}$ is continuous near the boundary of the ball $B^n_{1+\delta}$ and the argument given below can be applied to $\tilde{f}$ showing that the degree of $\tilde{f}$ is zero on the boundary of the ball $B^n_{1+\delta}$).

The mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy.$$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous up to the boundary: as $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f$ is continuous in an annulus near the boundary, $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$\left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}.$$ Since $$\operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all y}$$ we have $$\operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}.$$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|}$$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev spaces in mathematics. I, 185–222, Int. Math. Ser. (N. Y.), 8, Springer, New York, 2009.

[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

[3] R. Schoen, K. Uhlenbeck, Boundary regularity and the Dirichlet problem for harmonic maps. J. Differential Geom. 18 (1983), 253–268.

• I'm having trouble following part of the proof. How does the necessary condition that $f\vert_{\partial B^n}\in C^0$ get used in showing that $f_\epsilon$ is a continuous extension of $f\vert_{\partial B^n}$? Is it used to show that $f_\epsilon$ is continuous up to the boundary or to show that $f_\epsilon$ has the same trace as $f$? – Yasha Berchenko-Kogan Jul 13 '18 at 20:33
• @YashaBerchenko-Kogan I added some details showing continuity up to the boundary. You are right, my previous argument was kind of sloppy. If you have further questions, just ask. – Piotr Hajlasz Jul 14 '18 at 0:01
• Thanks! Continuity on a thin annulus resolves both questions. – Yasha Berchenko-Kogan Jul 16 '18 at 16:59

I believe I have an answer to my own question: the boundary map indeed must have degree zero, though I would still be curious about other proof methods.

The trick is that we can compute the degree of a map $g\colon\partial D^2\to S^1$ by computing $\int_{\partial D^2}\bar g\,dg$. This integral will be $2\pi i$ times the degree. But, moreover, this expression is well-defined for $g\in W^{1/2,2}(\partial D^2)$ because $\bar g\in W^{1/2,2}(\partial D^2)$ and $dg\in W^{-1/2,2}(\partial D^2)$. In other words, the notion of the degree of a map $\partial D^2\to S^1$ can be extended continuously from $C^\infty(\partial D^2;S^1)$ to $W^{1/2,2}(\partial D^2;S^1)$.

However, that's not the end of the story, because functions in $W^{1,2}(D^2;S^1)$ are limits of functions in $C^\infty(D^2;\mathbb C)$ that converge pointwise almost everywhere to $S^1$, and it is a nontrivial fact to show that they are actually limits of functions in $C^\infty(D^2;S^1)$, whose boundary restrictions indeed have degree zero. (Consider $e^{i\theta}$, which is in $W^{1,p}(D^2;S^1)$ for $p<2$, but to view it as a $W^{1,p}$-limit of smooth functions, we need to cut it off to zero near the origin.) Fortunately, Schoen and Uhlenbeck provide a proof in section 4 of Boundary regularity and the Dirichlet problem for harmonic maps.

In higher dimensions, it appears that this smooth approximation issue has also been resolved. See the introduction of this recent paper, for example. To generalize the above argument to higher dimensions, it remains to express the degree as an integral. Intuitively, we want to integrate the determinant of the Jacobian. For maps $g\colon\partial B^4\to S^3$, for example, the expression is \begin{equation*} \frac1{3!\cdot\operatorname{vol}(\partial B^4)}\int_{\partial B^4}g\wedge dg\wedge dg\wedge dg. \end{equation*} Explaining the notation, we view $g$ as a $\mathbb R^4$-valued zero-form on $\partial B^4$, and, correspondingly, $dg$ as a $\mathbb R^4$-valued one-form on $\partial B^4$. The $\wedge$ operation in this case means taking the usual wedge of forms on $\partial B^4$, along with taking the wedge in $\bigwedge^*\mathbb R^4$. (In particular, the operation is symmetric on $\mathbb R^4$-valued one-forms.) The resulting expression is a three-form on $\partial B^4$ with values in $\bigwedge^4\mathbb R^4\cong\mathbb R$.

Naively, because of borderline and negative regularity issues, this integral is not defined for $g\in W^{3/4,4}(\partial B^4;S^3)$. However, we can extend $g$ to a function in $W^{1,4}(B^4;\mathbb R^4)$ (not necessarily sphere-valued) so that the extension depends continuously on the boundary value. By Stokes' theorem, we see that \begin{equation*} \int_{\partial B^4}g\wedge dg\wedge dg\wedge dg=\int_{B^4}dg\wedge dg\wedge dg\wedge dg. \end{equation*} The right-hand side is indeed well-defined for $g\in W^{1,4}(B^4;\mathbb R^4)$.

See also the introduction of this Brezis and Nirenberg paper.

• The point about non-density of $C^{\infty}(D^2, S^1)$ in $W^{1,p}(D^2;S^1)$ is interesting. It seems that maybe the $C^{\infty}$ functions are dense in $W^{1,2}$, because of this degree argument. Intuitively, start with an approximating sequence $g_n$ of $C^{\infty}(D^2; \mathbb{C})$ functions and make them $S^1$ valued outside a disk of radius $(1/n)$. By the continuity of the integral you observed (applied to the small disk) you may argue that $g_n$ can be homotopic to a constant map in the small disk, and so they can be $S^1$ valued in the small disk as well. – Tim Carson Feb 11 '17 at 17:38
• Tom Mrowka pointed me to a Schoen and Uhlenbeck paper that discusses this density issue. See section 4. There's also a recent paper whose introduction seems to provide a good overview of what's currently known about smooth approximation for general Sobolev maps between manifolds. In particular, smooth maps $C^\infty(M;N)$ between compact manifolds are indeed dense in the borderline Sobolev spaces $W^{1,n}(M;N)$, where $n=\dim M$. I'll update my answer to reflect this. – Yasha Berchenko-Kogan Feb 14 '17 at 20:43

The paper Topology and Sobolev spaces by H. Brézis & Yanyan Li (Journal of Functional Analysis 183, 321-369 (2001)) considers the space $W^{1,p}(M,N)$ when $M$ and $N$ are compact connected oriented manifolds. It starts with the

Theorem 0.1. If $p\ge \dim M$, then $W^{1,p}(M,N)$ is path-connected if and only if $C^0(M,N)$ is path-connected.

They mention that the case $p=\dim M$ follows from ideas taken from Schoen & Uhlenbeck. In your situation, $M$ is $D^2$, $N$ is the unit circle, $p=2=\dim M$ and $C^0(M,N)=C^0(D^2,S^1)$ is path-connected because $M$ is contractile. Theorem $W^{1,2}(D^2,S^1)$ is path-connected. If there exists anything playing the role of a degree, the elements of this space must have degree zero.

• Since you were interested in the question you might look at my answer. I do not wee right away why the fact that $W^{1,2}(D^2,S^1)$ is path connected implies that the degree of the map on the boundary has to be zero. – Piotr Hajlasz May 6 '18 at 18:41

Once you take for granted your integral formula for the degree, you can proceed as follows: given $u\in W^{1,2}(D^2;S^1)$, take any Lebesgue point $x_0\in B_{1/2}$ for $u$. For any $t\in[0,1)$ let $$u_t(x):=u((2t\wedge 1)x_0+(1-t)x).$$ Notice that $t\mapsto u_t$ is a continuous path in $W^{1,2}(D^2;S^1)$ with $u_0=u$ and that $\lim_{t\to 1}u_t=c$ in $W^{1,2}(D^2;S^1)$, for some constant function $c\in W^{1,2}(D^2;S^1)$. So the degree of the trace of $u_t$ varies continuously from the original degree to $0$, giving the thesis.

Unfortunately, the only way I know to show that the degree is integer requires showing that $C^\infty(\partial D^2;S^1)\subset W^{1/2,2}(\partial D^2;S^1)$ is dense...

• Since you were interested in the question you might look at my answer. – Piotr Hajlasz May 6 '18 at 18:41