We know that if the curl vanishes: $$ \nabla \times \vec{v} = 0$$ then the field is irrotational and is conservative, but what about in higher dimensions than 1? The cross product is not defined there, so the above formula is no longer meaningful.
Are there similar conditions to determine if a vector field can be expressed as the gradient of a potential function?
Yes.
1. option: Another (equivalent) criterion: If the work done on a arbitrarily chosen closed path is zero, then the field is conservative. I.e:
$$ \oint \vec{F} \cdot \mathrm{d}\vec{r} = 0 $$
means $\vec{F}$ is conservative.
2. option: More general definition of curl
$$ \left(\mathrm{curl} \ \vec{F} \right)\cdot \vec{n} = \lim_{A \rightarrow 0} \left( \frac{1}{|A|} \oint_C \vec{F} \cdot \mathrm{d}\vec{r} \right) $$
where $A$ denotes area enclosed by the curve $C$ and $n$ is the normal vector to that area.
It is obvious that the both options are:
There are many conditions for identifying that a vector field $\vec v$ is conservative or not:
A conservative field vector is essentially irrotational.
Work done by a conservative vector field about any closed path $C$ is $0$.
A conservative vector field can always be represented as the gradient of a scalar potential.
Special test for 2-D vectors:
Let be $\vec v = P \hat i + Q \hat j$ a vector field on an open and simply-connected region $D$. Then if $P$ and $Q$ have continuous first order partial derivatives in $D$ and $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$
the vector field is conservative.
Further quoting from this link to clear the doubt about the curl of $2-D, 3-D$ vector fields,
If F is a three-dimensional vector field, $F:\mathbb{R}^3 \to \mathbb{R}^3$ , then we can derive another condition. This condition is based on the fact that a vector field $\vec F$ is conservative if and only if $\vec F=\nabla f$ for some potential function. We can calculate that the curl of a gradient is zero, curl $\nabla f=0$, for any twice continuously differentiable $f:\mathbb{R}^3 \to \mathbb{R}$. Therefore, if $\vec F$ is conservative, then its curl must be zero, as curl $F$ = curl $\nabla f=0$.
For a continuously differentiable two-dimensional vector field, $F:\mathbb{R}^2 \to \mathbb{R}^2$, we can similarly conclude that if the vector field is conservative, then the scalar curl must be zero, $$\frac{\partial F_2}{\partial x}−\frac{\partial F_1}{\partial y}=\frac{\partial f_2}{\partial x\partial y}−\frac{\partial f_2}{\partial y\partial x}=0$$ We have to be careful here. The valid statement is that if $\vec F$ is conservative, then its curl must be zero. Without additional conditions on the vector field, the converse may not be true, so we cannot conclude that $\vec F$ is conservative just from its curl being zero. There are path-dependent vector fields with zero curl. On the other hand, we can conclude that if the curl of $\vec F$ is non-zero, then F must be path-dependent.
