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It is known (see for instance Beauville - Determinantal hypersurfaces) that a generic homogeneous polynomial in $5$ variables of degree $5$ with complex coefficients can be written as the Pfaffian of a skew-symmetric $10 \times 10$ matrix with linear entries in the variables.

I was wondering if such a Pfaffian representation was known to exist for the Fermat quintic plynomial $F_5(X_1,X_2,X_3,X_4,X_5) = X_1^5 + X_2^5 + X_3^5 +X_4^5 + X_5^5$? If so, is there any reference which makes it explicit?

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Edit Jan 11, 2018: One can make a little bit of progress by pushing my initial idea (pair partitions plus round robin) "to second order", i.e., make use of a little bit of cancellation. See further below.


Edit Jan 10, 2018:

There is a flaw in the reasoning below as pointed out by Will.

One needs more conditions on the combinatorial design like the only way to produce a pair partition of $[10]$ only using pairs in $\cup_i P_i$ is to take one the $P_i$'s. I don't know if this extra requirement can be satisfied. Perhaps a round robin tournament scheduling algorithm might work.


Initial version (faulty!):

It follows from the existence of five set partitions $P_1,\ldots,P_5$ of the set $[10]:=\{1,2,\ldots,10\}$ into pairs such that these partitions are disjoint. Namely a pair $\{i,j\}\subset [10]$ with $i\neq j$ cannot be used by more than one partition.

Let us assume this existence. Then for each $P_i$ define its "standard permutation" as $\sigma_i$ given by listing the pairs by increasing order of its minimum and writing each pair in increasing order. Let $\epsilon_i$ be the sign of this standard permutation. We now define a tensor $T=(T_{i,j,k})_{(i,j,k)\in [10]\times[10]\times[5]}$ as follows. If a pair $\{i,j\}\in P_k$ with $i<j$, set $T_{i,j,k}=\epsilon_k$ and $T_{j,i,k}=-\epsilon_k$. Otherwise make all the other tensor entries zero. Define the matrix of linear forms $M_{i,j}(x)=\sum_{k=1}^{5}T_{i,j,k} x_k$. A moment of thought will show you that the Pfaffian of this matrix is a multiple of the Fermat quintic.

Fiddling around produces for instance this choice of partitions:

$$ P_1 =\{\ \{1,2\},\ \{3,4\},\ \{5,6\},\ \{7,8\},\ \{9,10\}\ \} $$ $$ P_2 =\{\ \{1,10\},\ \{2,3\},\ \{4,5\},\ \{6,7\},\ \{8,9\}\ \} $$ $$ P_3 =\{\ \{1,6\},\ \{2,7\},\ \{3,8\},\ \{4,9\},\ \{5,10\}\ \} $$ $$ P_4 =\{\ \{1,3\},\ \{2,10\},\ \{4,6\},\ \{5,8\},\ \{7,9\}\ \} $$ $$ P_5 =\{\ \{1,5\},\ \{2,6\},\ \{3,9\},\ \{4,7\},\ \{8,10\}\ \} $$


Edit Jan 11, 2018 (continued):

1) The Fermat quadratic $F=x_1^2+x_2^2$.

My construction works as is and one can take for the partitions $P_1$ and $P_2$ any two of the three pair partitions of $[4]$. To do this in a more principled way, let me modify the standard round robin scheduling algorithm in the even case. Instead of fixing $1$ and turning everybody else, I will turn $1$ too. This gives the two arrays $$ P_1=\left( \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array} \right)\ \ ,\ \ P_2=\left( \begin{array}{cc} 4 & 1 \\ 3 & 2 \end{array} \right)\ . $$ The pairs correspond to the columns of the arrays. The corresponding signs are $\epsilon_1=\epsilon_2=1$. This gives the matrix $$ M=\left( \begin{array}{cccc} 0 & x_2 & 0 & x_1 \\ -x_2 & 0 & x_1 & 0 \\ 0 & -x_1 & 0 & x_2 \\ -x_1 & 0 & -x_2 & 0 \end{array} \right) $$ with ${\rm Pf}(M)=F$.

2) The Fermat cubic $F=x_1^3+x_2^3+x_3^3$.

The modified round robin rule gives the three pair partitions $$ P_1=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 6 & 5 & 4 \end{array} \right)\ ,\ P_2=\left( \begin{array}{ccc} 6 & 1 & 2 \\ 5 & 4 & 3 \end{array} \right)\ ,\ P_3=\left( \begin{array}{ccc} 5 & 6 & 1 \\ 4 & 3 & 2 \end{array} \right)\ . $$ One can also use an array of pairs to see what happens: $$ \left( \begin{array}{ccc} (16) & (25) & (34) \\ (23) & (14) & (56) \\ (45) & (36) & (12) \end{array} \right) $$ The partitions $P_1$, $P_2$, $P_3$ correspond to the rows of this array and their signs are $\epsilon_1=\epsilon_2=\epsilon_3=1$.

A rather tedious inspection shows that the only pair partitions one can form using any pairs in this $3\times 3$ array are the previous ones and the three new ones corresponding to the columns of the array, i.e., $P_4=(16)(23)(45)$, $P_5=(25)(14)(36)$ and $P_6=(34)(56)(12)$. Remark that this looks nicer if one draws chord diagrams on a circle with three colors for $P_1$, $P_2$, $P_3$. The corresponding signs are $\epsilon_4=1$, $\epsilon_5=-1$ and $\epsilon_6=1$. Without any modification the Pfaffian should be $$ x_1^3+x_2^3+x_3^3+x_1 x_2 x_3-x_1 x_2 x_3+x_1 x_2 x_3 $$ where I emphasized the individual contributions of $P_1,\ldots,P_6$ on purpose. Now modify the weights of the pairs in the first row by factors $a,b,c$. The Pfaffian now becomes $$ abcx_1^3+x_2^3+x_3^3+ax_1 x_2 x_3-bx_1 x_2 x_3+cx_1 x_2 x_3 $$ so we get the Fermat cubic if we pick these weights so that $$ abc=1\ \ ,\ \ a-b+c=0 $$ For example we can take $a=\theta$, $b=1$, $c=\theta^5$ where $\theta=e^{\frac{i\pi}{3}}$.

More explicitly, the Fermat cubic is $F={\rm Pf}(M)$ with $$ M=\left( \begin{array}{cccccc} 0 & x_3 & 0 & x_2 & 0 & \theta x_1 \\ -x_3 & 0 & x_2 & 0 & x_1 & 0 \\ 0 & -x_2 & 0 & \theta^5 x_1 & 0 & x_3 \\ -x_2 & 0 & -\theta^5 x_1 & 0 & x_3 & 0 \\ 0 & -x_1 & 0 & -x_3 & 0 & x_2 \\ -\theta x_1 & 0 & -x_3 & 0 & -x_2 & 0 \end{array} \right)\ . $$

There is an obvious pattern here but I don't know if it works for $n=4$ or the OP's question, that is, $n=5$. In fact, I just realized (via performing the same permutation on rows and columns) these are determinantal cases as in the comment by Will.

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    $\begingroup$ This looks a rather interesting answer. I am not very familiar with this type of reasonning, but I am definitively very interested. Could you explain a bit more your sentence "A moment of thought..." $\endgroup$ – Libli Jan 10 '18 at 21:34
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    $\begingroup$ I see why the Fermat quintic terms appear, but why do the other terms vanish? For instance, it seems to me that the coefficient of $x_1^3 x_4 x_5$ in the Pfaffian of your matrix is nonzero, as the partition $\{ \{1,2\}, \{3,4\}, \{5,6\}, \{7,9\}, \{8,10\}\}$ contributes and no other partition does. $\endgroup$ – Will Sawin Jan 10 '18 at 21:49
  • $\begingroup$ @WillSawin: You are right. I went a bit too fast, so my argument as is does not work yet. $\endgroup$ – Abdelmalek Abdesselam Jan 10 '18 at 22:05
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    $\begingroup$ By a somewhat laborious search, I verified that this approach does not work for $n=3$. The combinatorial condition is as follows: each $\{a,b\}$ occurs at most once and there are no "4-cycles" of the form $\{a,b\}, \{b,c\}, \{c,d\}, \{d,a\}$. $\endgroup$ – Victor Protsak Jan 11 '18 at 0:10
  • $\begingroup$ @VictorProtsak: Thank you for your efforts. I did not quite get what $n$ is. $\endgroup$ – Abdelmalek Abdesselam Jan 11 '18 at 0:34
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This is not a complete answer, but I will give a concrete computation that shows that the Fermat quintic is in the adherence of the locus of Pfaffian quintics. Of course, this is a trivial consequence of Beauville's result (and Schreyer's computations with Macaulay2). But the proof I give is computer-free and might be of interest for concrete computations in this area. In fact, I will prove something more (which is not known to the best of my knowledge), the Fermat quintic in 6 variables is in the adherence of the Pfaffian locus.

The equivalence between $10 \times 10$ skew symmetric matrices and $5 \times 5$ hermitian matrices with quaternionic coefficients is well-known, so I might just be looking at $5 \times 5$ hermitian matrices with quaternionic coefficients. Let $t$ be a real paramater (which will go to $0$), $\sigma_1, \sigma_2, \sigma_3, \theta_1, \theta_2, \theta_3$ six quaternions and $l_1,l_2,l_3$ three real numbers. Let $M_t$ be the $5 \times 5$ hermitian matrices defined by:

$$M_t = \left( \begin{array}{ccccc} l_1 & 0 & \sigma_3 & 0 & \sigma_2 \\ 0 & l_2 & \sigma_1 & \theta_3 & 0 \\ \overline{\sigma_3} & \overline{\sigma_1} & 0 & \theta_2& 0 \\ 0 & \overline{\theta_3} & \overline{\theta_2} & 0 & \theta_1 \\ \overline{\sigma_2} & 0 &0& \overline{\theta_1} & l_3\\ \end{array} \right) $$

A tedious but simple computations shows that : $$ det(M_t) = l_1 \sigma_1 \overline{\sigma_1} \theta_1 \overline{\theta_1} + l_2 \sigma_2 \overline{\sigma_2} \theta_2 \overline{\theta_2} + l_3 \sigma_3 \overline{\sigma_3} \theta_3 \overline{\theta_3} + l_1l_3 Re(\sigma_1 \theta_2 \overline{\theta_3}) - l_2Re(\sigma_3 \theta_2 \theta_1 \overline{\sigma_2}) + Re(\sigma_3 \theta_3 \overline{\sigma_1} \theta_1 \overline{\sigma_2}) - \sigma_2 \overline{\sigma_2}Re(\sigma_1 \theta_2 \overline{\theta_3}) + l_2 \sigma_3 \overline{\sigma_3} \theta_1 \overline{\theta_1} - l_1l_2l_3 \overline{\theta_2} \theta_2.$$

If we let $l_1 = X_1+X_2$, $\sigma_1 = j\times(X_1 + \omega X_2)$, $\theta_1 = X_1 + \omega^2 X_2$, $l_2 = t^2(X_3+X_4)$, $\sigma_2 = \frac{1}{t} (X_3 + \omega X_4)$, $\theta_2 = X_3 + \omega^2 X_4$, $l_3 = X_5+X_6$, $\sigma_3 = X_5 + \omega X_6$, $\theta_3 = X_5 + \omega^2 X_6$, where $\omega = e^{i\frac{\pi}{5}}$ and $j$ is the "second quaternionic square root of $-1$".

Then we get:

$$ det(M_t) = X_1^5 + X_2^5 + X_3^5 + X_4^5 + X_5^5 + X_6^5 - t(X_3+X_4)\mathrm{Re}((X_5+\omega X_6)(X_3+ \omega^2 X_4)(X_1+\omega^2 X_2)(X_3+\overline{\omega}X_4)) + t^2(X_3+X_4)(X_5+\omega X_6)(X_5+\overline{\omega} X_6)(X_1+\omega^2 X_2)(X_1+\overline{\omega^2} X_2) -t^2(X_1+X_2)(X_3+X_4)(X_5+X_6)(X_3+ \omega^2X_4)(X_3+\overline{\omega^2}X_4).$$

The limit when $t \longrightarrow 0$ of $det(M_t)$ is $X_1^5+X_2^5+X_3^5+X_4^5+X_5^5+X_6^5$, which shows that the Fermat quintic in 6 variables is in the adherence of the Pfaffian locus.

Remark : The Fermat quintic in $7$ variables can't be in the adherence of the Pfaffian locus. Indeed, any six dimensional linear sections of the Pfaffian quintic in $\mathbb{P}^{44}$ is singular. But the Fermat quintic is always smooth, so it can't be in the adherence of a family of singular qintics. This shows that the "curve" of matrices found above is opitmal in some sense

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