Let $V$ a vector space of dimension $2$ over a field $k$ of characteristic different from $2$ and $3$. Let $S^{3}V$ the third symmetric power of $V$.
How to construct a symplectic form on $S^{3}V$ such that elements coming from the linear group of $V$ are similitudes for this form?
I believe it's some miraculous formula that I can't write down.

  • $\begingroup$ The linear group of $V$ contains many homotheties. Did you mean the special linear group of $V$? $\endgroup$ Jul 3, 2015 at 16:50
  • $\begingroup$ I edited my post ! $\endgroup$
    – Silam
    Jul 4, 2015 at 8:15

2 Answers 2


There's nothing miraculous about this: Here is an explicit formula: Let $x$ and $y$ be a basis for $V$. Then $A,B\in S^3(V)$ can be written in the form $$ A = a_{-3}\,x^3+3a_{-1}\,x^2y+3a_1\,xy^2+a_3\,y^3 \quad\text{and}\quad B = b_{-3}\,x^3+3b_{-1}\,x^2y+3b_1\,xy^2+b_3\,y^3 $$ where $a_i$ and $b_i$ are in $k$. Set $$ \langle A,B\rangle = a_{-3}b_3 - 3a_{-1}b_1 + 3a_1b_{-1} - a_3b_{-3}\,. $$ This anti-symmetric pairing is then preserved (up to a multiple) under basis change. I.e., if $\phi:V\to V$ is an isomorphism, it induces a map $\phi_3:S^3V\to S^3V$, and one has $$ \langle \phi_3(A),\phi_3(B)\rangle = \det(\phi)^{-3}\langle A,B\rangle. $$

Note that this generalizes to give a non-degenerate skew-symmetric pairing on $S^{2m+1}V$ when $V$ has dimension $2$ and the characteristic $p$ of $k$ does not divide any of the binomial coefficients ${2m{+}1}\choose{i}$ for $0\le i\le 2m{+}1$.

  • $\begingroup$ @WillSawin: Not sure what your comment means. When $m=1$, the only forbidden prime is $p=3$, so $p=2$ is OK. When $m=2$, $p=3$ is OK. When $m=3$, $p=2$ is OK. When $m=4$, $p=5$ is OK. $\endgroup$ Jul 4, 2015 at 16:41
  • $\begingroup$ Good point. I think the actual characterization is that for this to be nondegenerate, $2m+2$ must have a single nonvanishing digit in base $p$ notation - i.e. it is a power of $p$ times an integer less than $p$. $\endgroup$
    – Will Sawin
    Jul 4, 2015 at 17:35

Here is a construction that gives a way from where this formula comes from. I do not fully understand what the OP is trying to find in his last sentence. But this is what I think the question is about. Please downvote or comment if I misunderstood the intent of the question.

Let $V = k^{2^*}$. Denote by $X,Y$ the coordinate linear functionals for $V$. Thus, $$ V = \{ aX+bY ~ | ~ a,b\in k \} $$ The determinant $\omega:V\times V \to k$ given by, $$ \omega(aX+bY,cX+dY) = ad - bc$$ will turn $(V,\omega)$ into a symplectic space.

We can allow the group $G=\text{SL}_2(k)$ to act linearly on $V$ by inverse-transpose. A computation will show that this action preserves the symplectic structure. That is, $\omega(g.v,g.w) = \omega(v,w)$ where $g\in G$ and $v,w\in V$.

There is a natural action of $G$ on $\text{sym}^3 V$. Our goal is to construct a sympletic form on $\text{sym}^3 V$, denote again by $\omega$, such that the action of $G$ is still compatible with this new form.

The symmetric power $\text{sym}^3V$ is the space of all trilinear maps, $$ k^2 \times k^2 \times k^2 \to k $$ Denote by $x^3$ to be the following trilinear map, $$ (u,v,w) \mapsto X(u)X(v)X(w) $$ In a similar way we can construct $y^3:k^2 \times k^2 \times k^2 \to k$.

Denote by $3x^2y$ to be the following trilinear map, $$ (u,v,w) \mapsto X(u)X(v)Y(w) + X(u)Y(v)X(w) + Y(u)X(v)X(w) $$ It may be helpful to imagine $3x^2y$ as $xxy + xyx + yxx$.

In a similar manner one has the trilinear map $3xy^2:k^2 \times k^2 \times k^2 \to k$.

These four maps form a basis for $\text{sym}^3V$ and so, $$ \text{sym}^3V = \{ ax^2 + b(3x^2y) + c(3xy^2) + dy^3 ~ | ~ a,b,c,d\in k \} $$

We will use a small modification. Instead of the map $3x^2y$ we will denote by $x^2y$ as the map $\tfrac{1}{3}xxy + \tfrac{1}{3}xyx + \tfrac{1}{3}yxx$. This way we can use $x^2y$ and $yx^2$ in our basis of symmetric powers and restate, $$ \text{sym}^3V = \{ ax^2 + bx^2y + bxy^2 + dy^3 ~ | ~ a,b,c,d\in k \} $$

There is a trilinear universal map for the symmetric power, $$ V \times V \times V \to \text{sym}^3V $$ which is defined as follows, $$ (\ell_1,\ell_2,\ell_3) \mapsto \bigg( (u,v,w) \mapsto \frac{1}{3!}\sum_{\sigma\in S_3} \ell_{\sigma(1)} (u)\ell_{\sigma(2)} (v)\ell_{\sigma(3)}(w) \bigg) $$

It is easy to check that $(X,X,X)\mapsto x^3$, and likewise with all the other powers.

Given $(\ell_1,\ell_2,\ell_3) \in V\times V\times V$ we denote its image under this map by simply, $$ \ell_1.\ell_2.\ell_3$$ and this image is symmetric in each of the three variables.

If $G$ is a group which acts on $V$ then we can put an action of $G$ on $\text{sym}^3V$. Given $g\in G$ and $\ell_1.\ell_2.\ell_3 \in V$ the action is given by, $$ g(\ell_1.\ell_2.\ell_3) = (g\ell_1).(g\ell_2).(g\ell_3) $$ If the action of $G$ preserves $(V,\omega)$ i.e. $\omega(g\ell_1,g\ell_2) = \omega(\ell_1,\ell_2)$ then it is clear that the action of $G$ on $\text{sym}^3 V$ will preserve the symplectic form.

The space $V$ has a sympletic form $\omega$ given by the determinant, namely, $$ \omega(aX+bY,cX+dY) = ad - bc $$

We will use the sympletic space $(V,\omega)$ to construct a sympletic form on $\text{sym}^3V$ which we will denote again by $\omega$. This is accomplished by setting, $$ \omega( \ell_1.\ell_2.\ell_3 , \ell_1'.\ell_2'.\ell_3') = \frac{1}{3!} \sum_{\sigma\in S_3} \prod_{i=1}^3 \omega(\ell_i,\ell_{\sigma(i')}) $$

Since $\omega$ on $\text{sym}^3V$ is bilinear and alternating and since $x^3,x^2y,xy^2,y^3$ form a basis for $\text{sym}^3V$ it is sufficient to determine what $\omega$ does on each pair of these forms. That will give us an explicit rule to compute with $\omega$ using these coordinates.

We start with $\omega(x^3,x^2y)$, this is, $$ \omega(X.X.X,X.X.Y) = 0$$ this is because $\omega(X,X) = 0$, and any pairing of $X$ from $X.X.X$ will get paired with an $X$ from $X.X.Y$ no matter how these products are permuted. Thus, the end result is always zero.

In the same way $\omega(x^3,xy^2) = 0$. However, for $\omega(x^3,y^3)$ we get, $$ \omega(X.X.X,Y.Y.Y) = \frac{1}{3!} \sum_{\sigma\in S_3} \omega(X,Y)\omega(X,Y)\omega(X,Y) = 1$$

Next, $\omega(x^2y,y^3) = 0$ but for $\omega(x^2y,xy^2)$ we instead get, $$ \omega(X.X.Y,Y.Y.X) = \frac{1}{3!}\left( \omega(X,X)\omega(X,Y)\omega(Y,X) + \omega(X,X)\omega(X,Y)\omega(Y,X) \right) = -\tfrac{1}{3} $$

One can check $\omega(xy^2,x^2y) = \tfrac{1}{3}$.

More explicitly, if $P = ax^3 + bx^2y + cxy^2 + dy^3$ and $P' = a'x^3 + b'x^2y + c'xy^2 + d'y^3$, $$ \omega(P,P') = ad' - da' - \tfrac{1}{3}bc' + \tfrac{1}{3}cb' $$


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