Secant variety of the 2-uple embedding $\mathbb{P}^4\hookrightarrow\mathbb{P} ^ {14}$. Let $X\subset\mathbb{P} ^ {14} _{\mathbb{C}}$ be the image of the 2-uple embedding of $\mathbb{P}^4 _{\mathbb{C}}$ in $\mathbb{P}^{14} _{\mathbb{C}}$.
What is the secant variety $ Sec(X)=\overline{ \bigcup_{x_1,x_2\in X\atop x_1\neq x_2}\langle x_1, x_2\rangle }$ of $X$? What is its degree?
Thanks.
 A: As suggested by David Speyer, here I'm expanding my comment to his answer. 
J. Harris and L. Tu have  a paper ("ON SYMMETRIC AND SKEW-SYMMETRIC DETERMINANTAL VARIETIES", Topology, vol. 23, no. 1, 71-84, (1984)), where  they compute cohomology classes of degeneracy loci of symmetric and antisymmetric maps between locally free sheaves, providing an analogue of the classical Porteous formula.
As an application (Prop. 1.2), they derive  formulae for the degree of the varieties of symmetric/antisymmetric matrices of rank $\le r$.
A: I'm sorry I stumbled upon this old post and couldn't help but chime in. If you only care about the answer to the original question, you could use Macaulay2 and a simple script like this:
R = QQ[x_0..x_14];
S = QQ[y_0..y_4];
I = ker map(S,R,basis(2,S)); -- the ideal of the quadratic Veronese 
degree I -- its degree
codim I -- its codimension 

T = QQ[z_0..z_4];
ST = S**T;
bS = sub(basis(2,S),ST);
bT = sub(basis(2,T),ST);
J = ker map(ST,R,bS+bT); -- the ideal of the secant variety of the quadratic Veronese
degree J -- its degree
codim J -- its codimension

In a few seconds you get the same answer as David's.
A: This is the variety of $5 \times 5$ symmetric matrices of rank $\leq 2$. It is studied in Chapter 6.3 of Weyman's Cohomology of Vector Bundles and Syzygies. The defining equations are simply the $3 \times 3$ minors of the symmetric matrix.
One should be able to extract the degree from Weyman's Theorem 6.3.1. I'm having trouble following Weyman's notation, but I'll give it a shot later if someone else doesn't get it first.

OK, here is my attempted extraction. Let $Y$ be your secant variety. We will be computing the Hilbert function $H^0(Y, \mathcal{O}(t))$, and getting the degree from that.
The Lie Group $GL_5$ acts on $H^0(Y, \mathcal{O}(t))$, and Weyman's Theorem 6.3.1 is supposed to tell us how $H^0(Y, \mathcal{O}(t))$ decomposes into $GL_5$ irreducibles. I think my problem understanding Weyman's notation is that the bijection he is using between partitions and $GL_5$ representations is the transpose of the one I learned. Assuming I am right, his result specialized to our setting and expressed in the notation I learned is that
$$H^0(Y, \mathcal{O}(t)) \cong \bigoplus_{a+b=t,\ a \geq b} V_{2a, 2b}.$$
So the Hilbert function is
$$\sum_{a+b=t,\ a \geq b} \dim V_{2a, 2b}.$$
So, we need to compute the dimension of the representation $V_{2a,2b}$ of $GL_5$. This is the Schur function $s_{2a, 2b}$ evaluated at $(1,1,1,1,1,0,0,0,\ldots)$ where all the remaining variables are $0$. (That's $5$ ones because we're looking at $GL_5$.) We use the Jacobi-Trudi identity, equation A5 in Fulton and Harris.
$$s_{2a, 2b} = \det \begin{pmatrix} h_{2a} & h_{2a+1} \\ h_{2b-1} & h_{2b} \end{pmatrix}$$
Plugging in the $1$'s and $0$'s, we get
$$\dim V_{2a, 2b}= \det \begin{pmatrix} \binom{2a+4}{4} & \binom{2a+5}{4} \\ \binom{2b+3}{4} & \binom{2b+4}{4} \end{pmatrix}$$
$$ = (2a+4)(2a+3)(2a+2)(2b+3)(2b+2)(2b+1) \cdot 4 \cdot (2 a-2 b+1)/(4!)^2 .$$
$$ = \frac{8}{9} \left(a^3 b^3 (a-b) + \mbox{lower order terms} \right).$$
(Please double check this; it's very easy to make off-by-one errors in this sort of computation.)
So 
$$\sum_{a+b=t,\ a \geq b} \dim V_{2a, 2b} = \sum_{a+b=t,\ a \geq b} \frac{8}{9} a^3 b^3 (a-b) + O(t^6)$$
$$=\sum_{b=t}^{\lfloor t/2 \rfloor} \frac{8}{9} (t-b)^3 b^3 (t-2b) + O(t^6)$$
Approximating the sum by an integral only introduces lower order error terms. We have
$$\int_{x=0}^{t/2} \frac{8}{9} (t-x)^3 x^3 (t-2x) dx = \frac{t^8}{1152}$$
Putting it all together, the Hilbert function is
$$\dim H^0(Y, \mathcal{O}(t))  = \frac{t^8}{1152} + O(t^7)$$
We deduce that $Y$ is $8$ dimensional, and has degree 
$$8! \frac{1}{1152} = 35.$$

It is interesting that the dimension is $8$, as the naive guess would be $9$. ($4$ dimensions to choose a point on $\mathbb{P}^4$, another $4$ to choose another point, and $1$ dimension to choose a point on the line between them.) The fact that we fall short means that those points which are on secant lines are on a positive dimensional family of secant lines. Let's see why this is true.
As mentioned above, a generic point of $Y$ is some rank $2$ matrix. (There are also rank $1$ matrices, but those lie in a closed subvariety.) Without loss of generality, let's look at the matrix
$$X:=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0  \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$
Look at the $\mathbb{P}^2$ of matrices of the form 
$$\begin{pmatrix} a & b & 0 & 0 & 0 \\ b & c & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0  \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$
Inside this $\mathbb{P}^2$ is a conic of rank one matrices, given by the equation $ac=b^2$. For any point $(a:b:c)$ on this conic, the line through $(a:b:c)$ and $(1:0:1)$ meets the conic a second time. The line joining these two rank one matrices is a secant passing through $X$. So we see that any rank $2$ matrix lies on a positive dimensional pencil of secants.
This computation rules out some approaches to solving the problem. One idea I had was to consider the incidence variety $\mathcal{S}$ consisting of $(A,B, z)$ of triples where $A$ and $B$ are rank $1$ matrices and $z$ is a point on the line $\overline{AB}$. (There would be some issues when $A=B$, but let's ignore that for now.) So $\mathcal{S}$ has dimension $9$ and projection to $z$ gives a map $\mathcal{S} \to \mathbb{P}^{14}$ with image $Y$. If $Y$ were also $9$-dimensional, I could hope to do some computations in $H^{\ast}(\mathcal{S})$ and determine the pushforward of the fundamental class of $\mathcal{S}$ to $\mathbb{P}^{14}$. This would be the desired degree. But, since $\dim Y$ is $8$, this pushforward would just be zero.
