Convexity of a specific semialgebraic set I have an engineering problem that may be solved using semidefinite programming. I would like to know whether a given set is convex.
Let $m \in \mathbb{R}^+$ be a positive real scalar, $l \in \mathbb{R}^3$ be a real vector, and $L \succ 0$ be a size $3 \times 3$ real positive definite matrix. Does the following inequality
$$L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) \succ 0$$
where $S(\cdot)$ denotes the skew-symmetric matrix operator and $\succ 0$ denotes positive definiteness, define a convex set?

I.e.:
Being
$$m \in \mathbb{R}$$
$$l \equiv \left[l_x\ l_y\ l_z\right]^T$$

    $$
L \equiv \left[\begin{matrix}
      L_{xx} & L_{xy} & L_{xz} \\
      L_{xy} & L_{yy} & L_{yz} \\
      L_{xz} & L_{yz} & L_{zz} \\
    \end{matrix}\right]
$$

The variables $m,l_x,l_y,l_z,L_{xx},L_{xy},L_{xz},L_{yy},L_{yz},L_{zz}$ define a $\mathbb{R}^{10}$ space.
The constraints

    $$
\left\{
\begin{matrix}
  m &> 0 \\
  L &\succ 0\\
  L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) &\succ 0
\end{matrix}
\right.
$$

which, since $m>0$, are equivalent to

    $$
\left\{
\begin{matrix}
  m &> 0 \\
  L &\succ 0\\
  m L - S\left(l\right)^T S\left(l\right) &\succ 0
\end{matrix}
\right.
$$

define a semialgebraic set on the $\mathbb{R}^{10}$ variables space.
Here, $\succ 0$ means that the left argument is a positive-definite matrix, and,

    $$
S(x) = \left[\begin{smallmatrix}
  0 & -x_3 & x_2 \\
  x_3 & 0 & -x_1 \\
  -x_2 & x_1 & 0
\end{smallmatrix}\right]\quad\text{with}\quad
x = \left[x_1\ x_2\ x_3\right]^T
$$


I did some, manipulation and rewrote the last constraint as a polynomial inequalities system:
being

    $$
mI = m L - S\left(l\right)^T S\left(l\right) =
\left[\begin{smallmatrix}
  L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} & L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1xz} m_{1} + l_{1x} l_{1z} \\
  L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1yy} m_{1} - l_{1x}^{2} - l_{1z}^{2} & L_{1yz} m_{1} + l_{1y} l_{1z} \\
  L_{1xz} m_{1} + l_{1x} l_{1z} & L_{1yz} m_{1} + l_{1y} l_{1z} & L_{1zz} m_{1} - l_{1x}^{2} - l_{1y}^{2}
\end{smallmatrix}\right]
$$

then, through Sylvester's criterion,
  
    $$
mI \succ 0 \Leftrightarrow
\left\{
\begin{matrix}
\det\left(mI_{1,1}\right) = L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} &> 0\\
\det\left(mI_{1:2,1:2}\right) &> 0\\
  \det\left(mI\right) &>0
\end{matrix}
\right.
$$

It would be sufficient that the polynomials were concave to guarantee set convexity, however they are not concave.
Although not being concave, it does not imply that set is not convex; for example, the first polynomial $L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2}$ is not concave itself but defines a convex set if constraint $m>0$ is taken into account.
(This representation also gave me some suspicions that maybe the $L \succ 0$ constraint is implicit on the other.)
I also tried to write the set as a linear matrix inequality (LMI), but I couldn't (my knowledge in this area is really short). 

Update:
I was able to check that this set is close under positive scalar multiplication, since
$$ (\gamma\ m) (\gamma\ L) - S\left(\gamma\ l\right)^T S\left(\gamma\ l\right) = \gamma^2 \ \left( m L - S\left(l\right)^T S\left(l\right)\right) \succ 0 \quad \text{for} \quad \gamma > 0$$
then it is a cone. If one can prove the set is close under addition then it will be proven to be a convex cone.


Now, the questions are:
Which methods can I use to check if the defined set is convex?
If so, is it possible to represent it as an LMI?

 A: Your set is indeed a convex cone.
Since it is a cone, it suffices to show that the $m=1$ section is convex. 
But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.
Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.
EDIT : in fact any set of $(x,y)$ defined by an inequality $S(y)-A(x)^*A(x) \succ 0$, with $S(y)$ $n\times n$ symmetric and linear in $y$, and $A(x)$ $n\times d$ and linear in $x$, is convex and moreover defined by a Linear Matrix Inequality.
Indeed this is equivalent to 
$$\left(\begin{matrix}
          S(y) & A(x)^*  \\\
          A(x) & I  \end{matrix}\right) \succ 0$$
as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.
