Geometry of Elementary Symmetric Polynomials 
The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the identity
  $$
    \prod _{j=1}^{n}(\lambda -X_{j})=\lambda ^{n}-e_{1}(X_{1},\ldots ,X_{n})\lambda ^{n-1}+e_{2}(X_{1},\ldots ,X_{n})\lambda ^{n-2}+\cdots +(-1)^{n}e_{n}(X_{1},\ldots ,X_{n}). 
$$
  For $n = 3$:
  $$
{\begin{aligned}
e_{1}(X_{1},X_{2},X_{3})&=X_{1}+X_{2}+X_{3},\\
e_{2}(X_{1},X_{2},X_{3})&=X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3},\\
e_{3}(X_{1},X_{2},X_{3})&=X_{1}X_{2}X_{3}.\\\
\end{aligned}} 
$$

In the example given, we get a plane for $e_{1}(X_{1},X_{2},X_{3})=c_1$ and two-sheeted hyperboloid for  $e_{2}(X_{1},X_{2},X_{3})=c_2$, which gives a cone if $c_2=0$.
Is there a general description of the geometry of $e_k(X_1,\dots,X_n)=c_k$?
 A: Here's a partial answer; we'll treat $k = 1, 2, n$ and $(n, k) = (4, 3)$.
Note, by the way, that if $c_k = 0$, the resulting equation $e_k(X_1, \ldots, X_n) = 0$ is homogeneous, so by projectivizing we may regard the solution set as a codimension $1$ projective variety in $\Bbb P^{n - 1} = \Bbb P(\Bbb F^n)$.
Now, of course, $$e_1(X_1, \ldots, X_n) = X_1 + \cdots + X_n = c$$ defines a hyperplane, namely, the one through $(c, 0, \ldots, 0)$ with normal vector $${\bf U} := \pmatrix{1\\ \vdots\\1}.$$
For any $n$, we can regard $$e_2(X_1, \ldots, X_n) = \sum_{i < j} X_i X_j$$ as a quadratic form on $\Bbb R^n$ endowed with coordinates ${\bf X} := (X_a)$, namely the one with associated matrix
$$[e_2] := \frac{1}{2}\,
\pmatrix{
0 & 1 & 1 & \cdots & 1\\
1 & 0 & 1 & \cdots & 1\\
1 & 1 & 0 & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 0} = \frac{1}{2}({\bf U}^T {\bf U} - I_n)$$
with respect to the standard basis. One can show that $[e_2]$ has nonzero determinant (for example, since it is a rank-one update of the invertible matrix $-\frac{1}{2}I_n$, we can use the Sherman-Morrison Formula), so the quadratic form is nondegenerate (at least for $n > 1$; for $n = 1$, $e_2(X_1) = 0$, and we henceforth disregard this case). It's easy to see $[e_2]$ has $1$ positive eigenvalue and $n - 1$ negative eigenvalues, so $e_2$ has signature $(1, n - 1)$. We can conclude the following about the level set
$$
\Sigma_c := \{e_2({\bf X}) = c\}, \qquad c \in \Bbb R .
$$


*

*For $c > 0$, $\Sigma_c$ is a nondegenerate $2$-sheeted quadric hypersurface. (The two sheets are separated by the hyperplane $e_1({\bf X}) = \{{\bf U} \cdot {\bf X} = 1 \}$.)

*For $c = 0$, $\Sigma_0$ is a nondegenerate cone; its projectivization is an $(n - 2)$-sphere in $\Bbb R \Bbb P^{n - 1}$.

*For $c < 0$, $\Sigma_c$ is a nondegenerate $1$-sheeted quadric hypersurface for $n > 2$ (but again is $2$-sheeted for $n = 2$, in which case it is simply a hyperbola).


When $k = n$, the variety $\{e_n({\bf X}) = 0\}$ is the union of the coordinate hyperplanes $\{X_a = 0\}$, $a = 1, \ldots, n$.
The varieties $\{e_k({\bf X}) = c\}$ not covered by the above cases are generally less familiar, but at least some of them were studied classically and even have specific names. For example, in the simplest remaining case, $n = 4$, $k = 3$, i.e., the variety $\{e_3(X_1, X_2, X_3, X_4) = 0\}$ is sometimes called Cayley's Surface (or more precisely, its projectivization in $\Bbb R \Bbb P^3$ is).
