Secant varieties of curves in $\mathbb{P}^4$ My question is motivated by the following simple observations. By a standard dimensions count in $\mathbb{P}^4$ there should not exist neither an hypersurface of degree $3$ with multiplicity $2$ in seven general points, nor an hypersuperface of degree $5$ with multiplicity $3$ in eight general points. On the other hand:


*

*through seven general points in $\mathbb{P}^4$ there is a rational normal curve $C$ of degree $4$ and its secant variety $Sec_2(C)$ is an hypersurface of degree $3$ with multiplicity $2$ along $C$ and in particular in the seven points;

*through eight general points in $\mathbb{P}^4$ there is an elliptic normal curve $E$ and its secant variety $Sec_2(E)$ is an hypersurface of degree $5$ with multiplicity $3$ along $E$ and in particular in the eight points.


My question is: does there exist in $\mathbb{P}^4$ an hypersurface of degree $6$ and with multiplicity $4$ in eight general points? Perhaps the secant variety of some special curve in $\mathbb{P}^4$.
 A: I think the answer is no. Here is an argument.
Degenerate your $8$ points so that there are $2$ $5$-tuples that lie in $3$-planes $A,B$. The degree $3$ rational curves through the first $5$-tuple intersect the hypersurface with multiplicity at least $5*4>3*6,$ so they lie in the hypersurface. But such degree $3$ rational curves span $A$ (as there is a rational normal curve in $\mathbb{P}^3$ through any $6$ points), so the hypersurface must contain $A$. Similarly, the hypersurface must contain $B$. 
Now residuating, we get a degree $4$ hypersurface with multiplicity $2$ at the $2$ common points of the $5$-tuples and multiplicity $3$ at the other points. As $3*3+2*2>3*4,$ the same argument shows that this residual hypersurface also contains $A$ and $B$.
Residuating again, we get a degree $2$ hypersurface with multiplicity $2$ at $6$ points (which can be chosen to be in general position). But a degree $2$ hypersurface is a cone over the span of its singularities, so this degree $2$ hypersurface cannot exist, contradiction.
A: The answer is no. You can see this by iterating a standar Cremona transformation. 
Let $p_1,...,p_{n+1}\in\mathbb{P}^n$ be general points. We may assume
$$p_1 = [1:0:...:0],...,p_{n+1} = [0:...:0:1].$$
We consider the standard Cremona transformation:
$$
\begin{array}{ccc}
\psi:\mathbb{P}^n & \dashrightarrow & \mathbb{P}^n\\
\left[x_0:...:x_n\right] & \longmapsto & [\frac{1}{x_0}:...:\frac{1}{x_n}]
\end{array}
$$
Note that $\psi\circ \psi = Id_{\mathbb{P}^n}$, and $\psi^{-1} = \psi$. Let $H_1,...,H_{n+1}$ be the coordinate hyperplanes of $\mathbb{P}^n$. Then $\psi$ is not defined on the locus
$$\bigcup_{1\leq i< j\leq n+1}H_i\cap H_j.$$
Furthermore, $\psi$ is an isomorphism off of the union
$$\bigcup_{1\leq i\leq n+1}H_i.$$
Let $X_{n+1}^n$ be the blow-up of $\mathbb{P}^n$ is the $n+1$ base points of the Cremona. Now, $\psi$ induces a birational transformation $\widetilde{\psi}:X_{n+1}^n\dashrightarrow X_{n+1}^n$.
Note that, since $\psi$ contracts the hyperplane $H_i$ passing spanned by the $n$ points $p_1,...,\hat{p}_i,...,p_{n+1}$ to the point $p_i$, the map $\widetilde{\psi}$ maps the strict transform of $H_i$ onto the exceptional divisor $E_i$. Therefore $\widetilde{\psi}$ is an isomorphism in codimension one. Indeed, it is a composition of flops. In particular $\widetilde{\psi}$ induces an isomorphism $Pic(X_{n+1}^n)\rightarrow Pic(X_{n+1}^n)$.
Now, the linear system on $\mathbb{P}^n$ associated to the standard Cremona $\psi$ is 
$$\mathcal{H} = \mathcal{O}_{\mathbb{P}^n}(n)\otimes \mathcal{I}_{(n-1)(p_1+...+p_{n+1})},$$
that is $\mathcal{H}$ is the linear system of hypersurfaces in $\mathbb{P}^n$ of degree $n$ having points of multiplicity at least $n-1$ in $p_1,...,p_{n+1}$. Therefore, the inverse image of a general hyperplane of $\mathbb{P}^n$ via $\psi$ is an hypersurface of degree $n$ with points of multiplicity $n-1$ in $p_1,...,p_{n+1}$, and
$$\widetilde{\psi}^{*}H = nH-(n-1)(E_1+...+E_{n+1}).$$ 
Furthermore, since $\psi$ contracts the hyperplane $H_i$ passing spanned by the $n$ points $p_1,...,\hat{p}_i,...,p_{n+1}$ to the point $p_i$ we have
$$\widetilde{\psi}^{*}E_i = H-E_1-...-\hat{E}_i-...-E_{n+1}.$$ 
Let $D\subset\mathbb{P}^n$ be an hypersurface of degree $d$ having points of multiplicities $m_1,...,m_{n+1}$ in $p_1,...,p_{n+1}$, and let $\psi:\mathbb{P}^{n}\dashrightarrow\mathbb{P}^n$ be the standard Cremona of $\mathbb{P}^n$. Then 
$$\deg(\psi(D)) = dn-\sum_{i=1}^{n+1}m_i$$
and
$$mult_{p_i}\psi(D) = d(n-1)-\sum_{j\neq i}m_j$$
for any $i = 1,...,n+1$.
proof: Let $X_{n+1}^n = Bl_{p_1,...,p_{n+1}}\mathbb{P}^n$, and $\widetilde{\psi}:X_{n+1}^n\dashrightarrow X_{n+1}^n$ be the birational map induced by $\psi$. The strict transform of $D$ in $X_{n+1}^n\dashrightarrow X_{n+1}^n$ can be written as $\widetilde{D} \cong dH-\sum_{i=1}^{n+1}m_iE_i$.\
Now, since $\widetilde{\psi}_{*}H = nH-\sum_{i=1}^{n+1}(n-1)E_i$, and $\widetilde{\psi}_{*}E_i = H-\sum_{j\neq i}E_i$ we get
$$
\begin{array}{ll}
\widetilde{\psi}_{*}D = & d(nH-\sum_{i=1}^{n+1}E_i)-\sum_{i=1}^{n+1}m_i(H-\sum_{j\neq i}E_j)=\\
 & dnH-d\sum_{i=1}^{n+1}(n-1)E_i-\sum_{i=1}^{n+1}H+\sum_{i=1}^{n+1}m_i\sum_{j\neq i}E_j =\\ 
 & (dn-\sum_{i=1}^{n+1}m_i)H-\sum_{i=1}^{n+1}(d(n-1)-\sum_{j\neq i}m_j)E_j.
\end{array} 
$$
Now, assume there exists an hypersurface $X$ of degree $d = 6$ in $\mathbb{P}^4$ with eight points of multiplicity four. Consider a standard Cremona $f_1$ centred in five of this eight points. Let $X_1 = f(X)$. Then $X_1$ is an hypersurface of degree $d_1 = 4$ with five points of multiplicity $2$ and three points of multiplicity $4$. Now consider another standard Cremona $f_2$ centred in three points of multiplicity $2$ of $X_1$, and two points of multiplicity $4$ of $X_1$. Then $X_2 =f_2(X_1)$ is an hypersurface of degree $4\cdot 4-4-4-2-2-2 = 2$ with a point of multiplicity $4$. A contradiction.
