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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 $\mathbb{P}^4$.

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  • $\begingroup$ An observation: If one exists, it must contain all irreducible degree 5 rational curves through those 8 points. Now if the 8 points can be degenerated so that this space is larger than usual, then you can probably prove nonexistence... $\endgroup$
    – dhy
    Commented May 23, 2015 at 15:53
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    $\begingroup$ dhy gives a reasonable sounding argument that the answer is no. Just a remark: this sort of thing is easy to check in Macaulay2 or similar, since you can just generate a bunch of random points in $\mathbb P^n$ and compute the Hilbert function. In this case I found there is no such surface for some random set of points, which implies there are no sections for general points either. Of course this isn't very satisfying, but if you're going to want to check a lot of things like this it's worth learning how (sooner rather than later! I should've learned it long before I did.) $\endgroup$
    – user47305
    Commented May 24, 2015 at 2:25

2 Answers 2

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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.

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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.

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