Condition for an expression to be a total differential

Question

I have fully understood the concept and formulae around total differentials of multivariate functions. What is the condition however for an expression of differentials to be the total differential of a function? More specifically, what is the condition so that $ \Phi(dx, dy, dz) = df $ where $\Phi$ is an expression and $f:\mathbb{R^3} \to \mathbb{R}$?

Answer 1

Such an "expression" (really, a differential form) $\Phi(dx,dy,dz)$ is said to be exact if there exists some $f \in C^\infty(\mathbb{R}^3)$ such that $\Phi = df$. This is what you're looking for.

A necessary condition for a form to be exact is that it has to be closed, i.e. $d\Phi = 0$ itself. This is because $d(df)$ is always zero, a result following from the symmetry of partial derivatives (a good exercise). As it turns out, by the Poincaré lemma, this is also sufficient: as soon as $d\Phi = 0$, there exists $f$ such that $\Phi = df$. This is something very special about $\mathbb{R}^n$, or more generally contractible open subsets of $\mathbb{R}^n$.

For more complicated manifolds (or, if you haven't learned about manifolds yet, you can think about open subsets of $\mathbb{R}^n$), this is no longer true. An exact form (i.e. a form $\Phi$ of the type $df$) is necessary closed (i.e. $d\Phi = 0$), but the converse is not necessarily true: there may be closed 1-forms that aren't exact. An example is given by the form $(-ydx+xdy)/(x^2+y^2)$ on $\mathbb{R}^2 \setminus \{0\}$). This is encoded by de Rham cohomology.