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If $f$ is a continuous function from the unit circle to the unit circle, then $\deg(f)$ is the number of times $f$ wraps the target circle when its argument wraps the source circle once. It is always an integer number, since the travel always starts and ends at the same point.

If $g$ is a continuous function from an interval $[0,1]$ to the unit circle $S^1$, then the number of wraps can be non-integer, since the travel around the target circle can end at a different point then where it started. So the "degree" of $g$ can be fractional.

Is there any meaning to a fractional degree in two or more dimensions?

Specifically, if $h$ is a function from a $d$-dimensional square $[0,1]^d$ to a $d$-dimensional sphere $S^d$, is there a meaningful way to define the degree of $h$?

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    Rather than saying the "degree" of $g$ can be fractional, it's better to say that it can be a real number. – 2017-01-06

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For continuously differentiable maps this is straightforward.

For the case of $g : [0,1] \to S^1$, the "degree" is a real number given by the integral: $$\frac{1}{2\pi} \int_{[0,1]} \omega(g'(x)) \, dx $$ where $g'(x)$ is the tangent vector at $g(t) \in S^1$, and $\omega$ is the 1-form that eats a tangent vector and spits out its signed length, with a $+$ sign for counterclockwise pointing vectors and a $-$ sign for clockwise pointing vectors.

For the case of $h : [0,1]^d \to S^d$ you can do something similar letting $\omega$ be the area form $dA$ on $S^d$, which eats a $d$-tuple of vectors tangent to each point and spits out the signed area of the parallelogram spanned by those vectors, a $+$ sign if they satisfy the right hand rule with respect to the outward pointing normal on $S^d$, and a $-$ sign otherwise. Then you can define the degree to be $$\frac{1}{4\pi} \int_{[0,1]^d} \omega(Dh_x(e_1),...,Dh_x(e_d)) \, dx_1 \ldots dx_d $$ where $e_1,...,e_d$ are the standard orthogonal unit basis vectors at each point of $[0,1]^d$.

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    What does $Dg_x$ mean? Should it be $Dh_x$ and mean the derivative of the vector-function $h$ as a function of the vector $x$? – 2017-01-07
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    Also, is there a term for the latter expression, like "the degree of $h$" or something similar? – 2017-01-07
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    Oops, yes, I fixed $Dg_x$ to be $Dh_x$ which is as you say. – 2017-01-07
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    Using the term "degree" in the context of $g$ and $h$ is not standard. But just as the formula for $g$ does indeed become the degree if $g(0)=g(1)$, something similar happens with $h$. If $h$ is constant on the boundary of the $d$-cube $[0,1]^d$ then it induces, via the quotient map that crushes the boundary to a point, a map $S^d \mapsto S^d$ and in that setting yes, this integral is called the degree. – 2017-01-07