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.



Secant variety of the 2-uple embedding $\mathbb{P}^4\hookrightarrow\mathbb{P} ^ {14}$.
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](http://books.google.com/books?id=t_jdqfMMtnYC&lpg=PA229&ots=M21MRTokWP&dq=%2522rank%2520variety%2522%2520symmetric&pg=PA175#v=onepage&q=symmetric%2520matrices&f=false). 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.




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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](http://books.google.com/books?id=6GUH8ARxhp8C&lpg=PP1&pg=PA455#v=onepage&q&f=false).
$$s\_{2a, 2b} = \det \begin{pmatrix} h\_{2a} & h\_{2a+1} \\ h\_{2b-1} & h\_{2b} \end{pmatrix}$$
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 
$$8! \frac{1}{1152} = 35.$$




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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.


