A Question about Non-Conservative Vector Fields

Question

In my multivariable calculus class, we spent some time discussing the vector field that was the gradient of arctan(y/x). This field was shown to be non-conservative in closed regions which enclosed the origin. Our instructor also hinted at the idea that any non-conservative vector field could be described as the sum of a conservative gradient field and a multiple of (-y,x)/(x^2+y^2), which is the gradient of the function mentioned above. This idea reminded me of algebra, where, in a linear map, the image of any set and the image of sums of members of the set and the kernel are identical. My question now is, is there any sort of relationship between these two notions of kernel in an algebraic structure and conservative vector fields? Furthermore, does it make sense to talk about the dimension of the gradient operator, as a sort of local linear map, in this context?

Answer 1

Let me give you some more details. When I wrote my comment I was in a hurry and I expected you would be confused since de Rham cohomology uses some really heavy tools. Here's a taste for what's happening.

Your professor was talking about the following theorem.

Every smooth irrotational vector field defined on $\mathbb R^2-\{(0,0)\}$ is the sum of a conservative vector field and a scalar multiple of the vector field $(-y,x)/(x^2+y^2)$.

(By "smooth" I mean "derivatives of every order exist and are continuous everywhere". "Irrotational" means the curl is zero -- I'll elaborate in a second.) How can we relate this theorem to algebra? Let $V$ denote the vector space of smooth functions defined on $\mathbb R^2-\{(0,0)\}$ and let $W$ denote the vector space of smooth vector fields on $\mathbb R^2-\{(0,0)\}$. (So the "vectors" in $V$ are functions and the "vectors" in $W$ are vector fields -- the idea's a little weird if you're not used to it. These are huge spaces.)

The gradient operator $\nabla$ is a linear map from $V$ to $W$. That is, if $f$ is a function then $\nabla f$ is a vector field. Similarly, we can think of the curl operator as a linear map from $W$ to $V$ if we ignore the final vector value of the curl and write $$ \operatorname{curl}(f_1,f_2) = \frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2}. $$ Under the definitions I made, the conservative vector fields are exactly the elements of $W$ in the image of $\nabla$ (a conservative vector field is the gradient of some function, after all), and the irrotational vector fields are exactly the elements of $\ker(\operatorname{curl})$. Let me rephrase the theorem in this way:

The quotient space $\ker(\operatorname{curl})/\operatorname{Im}(\nabla)$ has basis $\{(-y,x)/(x^2+y^2)\}$ and is therefore one-dimensional.

Now I'm going to say something that hints at the depth you can reach by pursuing these ideas further. The reason that the vector space $\ker(\operatorname{curl})/\operatorname{Im}(\nabla)$ (called the first de Rham cohomology group) has dimension one is because the space $\mathbb R^2-\{(0,0)\}$ has one hole, namely the missing point $(0,0)$.