On Some Subspaces of the Exterior Algebra of a Simple Lie Algebra

Abstract

In this article, we are interested in some subspaces of the exterior algebra of a simple Lie algebra $\mathfrak {g}$. In particular, we prove that some graded subspaces of degree d generate the ${\mathfrak {g}}$-module $\bigwedge ^{d}{\mathfrak {g}}$ for some integers d.

Appendix A: Some Remarks on Root Systems

Let β be in Π and X := Π ∖{β}. Set ${\mathfrak p}_{\mathrm {u}} := {\mathfrak p}_{\mathrm {u},X}$ and $d := \dim \hskip .125em {\mathfrak p}_{\mathrm {u},X}$. The goal of the section is the following proposition:

Proposition A.1

• (i) Suppose Π of type A_ℓ and X not connected. Then β = β_s+ 1 for some s in {1,…, ℓ − 2},

$$ \mathrm{m} =2, \quad n_{1} = \frac{s(s+1)}{2}, \quad n_{2} = \frac{(\ell-s-1)(\ell-s)}{2} . $$

Moreover, if 2d + n₁ ≤ n then

$$ \ell \geq 6, \quad s \leq \frac{1}{6}\left( 2\ell -3 - \sqrt{ 4\ell^{2} + 12 \ell + 9}\right), \quad 2d+n_{2} > n . $$

• (ii) Suppose Π of type B_ℓ or C_ℓ and X not connected. Then β = β_s+ 1 for some s in {1,…, ℓ − 2},

$$ \mathrm{m} =2, \quad n_{1} = \frac{s(s+1)}{2}, \quad n_{2} = (\ell-s-1)^{2} . $$

Moreover, if 2d + n₁ ≤ n then

$$ \ell \geq 7, \quad s \leq \frac{1}{10}\left( 8\ell -9 - \sqrt{ 24\ell^{2} + 16 \ell + 1}\right), \quad 2d+n_{2} > n . $$

• (iii) Suppose Π of type D_ℓ. If β = β_ℓ− 2 then 2d > n. If β is different from β_ℓ− 2 and X is not connected, then

$$ \mathrm{m} =2, \quad n_{1} = \frac{s(s+1)}{2}, \quad n_{2} = (\ell-s-1)^{2} . $$

Moreover, if 2d + n₁ ≤ n then

$$ \ell \geq 8, \quad s \leq \frac{1}{10}\left( 8\ell -13 - \sqrt{ 24\ell^{2} - 8\ell + 9}\right), \quad 2d+n_{2} > n . $$

• (iv) Suppose that Π is exceptional. If 2d ≤ n then X is connected.

We prove the proposition case by case. So, in the classical case, we suppose ℓ ≥ 3 and X not connected.

1.1 A.1 Type A_ℓ

As X is not connected, m = 2 and β = β_s+ 1 for some s in {1,…,ℓ − 2}. Then

$$ \begin{array}{@{}rcl@{}} n_{1} = \frac{s(s+1)}{2}, \quad n_{2}& =& \frac{(\ell-s-1)(\ell-s)}{2}, \quad d = n -n_{1}-n_{2},\\ n - 2d - n_{1} &=& \frac{1}{2}\left( 3s^{2} + (-4\ell + 3)s + \ell^{2}- 3\ell\right). \end{array} $$

If n − 2d − n₁ ≥ 0 then

$$ s \leq \frac{1}{6}\left( 4\ell -3 - \sqrt{4\ell^{2} + 12\ell + 9}\right) \quad \text{or} \quad s \geq \frac{1}{6}\left( 4\ell -3 + \sqrt{4\ell^{2} + 12\ell + 9}\right) . $$

As s ≥ 1, the first inequality is possible only if ℓ ≥ 6. The second inequality is impossible since its right hand side is bigger than ℓ − 2 and s is at most ℓ − 2.

By the above equalities,

$$ n -2d - n_{2} = 3s^{2} + (-2\ell + 3)s - 2\ell. $$

If the left hand side is nonnegative then

$$ s \leq \frac{1}{6}\left( 2\ell -3 - \sqrt{4\ell^{2} + 12\ell + 9}\right) \quad \text{or} \quad s \geq \frac{1}{6}\left( 2\ell -3 + \sqrt{4\ell^{2} + 12\ell + 9}\right) . $$

The first inequality is impossible since its right hand side is negative. The second inequality is possible only if ℓ ≥ 7 since s ≤ ℓ − 2. Moreover, it is not possible to have n ≥ 2d + n₁ and n ≥ 2d + n₂ since

$$ \frac{1}{6}\left( 2\ell -3 + \sqrt{4\ell^{2} + 12\ell + 9}\right) > \frac{1}{6}\left( 4\ell -3 - \sqrt{4\ell^{2} + 12\ell + 9}\right), $$

whence Assertion (i) of Proposition A.1.

1.2 A.2 Type B_ℓ or C_ℓ

As X is not connected, m = 2 and β = β_s+ 1 for some s in {1,…,ℓ − 2}. Then

$$ \begin{array}{@{}rcl@{}} n_{1} = \frac{s(s+1)}{2}, \quad n_{2} &=& (\ell-s-1)^{2}, \quad d = n -n_{1}-n_{2},\\ n - 2d - n_{1} &=& \frac{1}{2}\left( 5s^{2} + (-8\ell + 9)s + 2\ell^{2}- 8\ell + 4\right). \end{array} $$

If n − 2d − n₁ ≥ 0 then

$$ s \leq \frac{1}{10}\left( 8\ell -9 - \sqrt{24\ell^{2} + 16\ell + 1}\right) \quad \text{or} \quad s \geq \frac{1}{10}\left( 8\ell -9 + \sqrt{24\ell^{2} + 16\ell + 1}\right) . $$

As s ≥ 1, the first inequality is possible only if ℓ ≥ 7. The second inequality is impossible since its right hand side is bigger than ℓ − 2 and s is at most ℓ − 2.

By the above equalities,

$$ n -2d - n_{2} = 2s^{2} + (-2\ell + 5)s - 4\ell+4. $$

If the left hand side is nonnegative then

$$ s \leq \frac{1}{4}\left( 2\ell -5 - \sqrt{4\ell^{2} + 12\ell - 7}\right) \quad \text{or} \quad s \geq \frac{1}{4}\left( 2\ell -5 + \sqrt{4\ell^{2} + 12\ell - 9}\right) . $$

The first inequality is impossible since its right hand side is negative. The second inequality is impossible since

$$s\leq \ell -2 \quad \text{and} \quad \frac{1}{4}\left( 2\ell -5 + \sqrt{4\ell^{2} + 12\ell - 9}\right) > \ell -2, $$

whence Assertion (ii) of Proposition A.1.

1.3 A.3 Type D_ℓ

As X is not connected, β is different from β₁, β_ℓ− 1, β_ℓ. If β = β_ℓ− 2 then X has three connected components and

$$ d = \ell(\ell -1) - 2 - \frac{1}{2}(\ell-3)(\ell -2) . $$

In this case n < 2d. Suppose ℓ ≥ 5 and β = β_s+ 1 for some s in {1,…,ℓ − 4}. Then

$$ \begin{array}{@{}rcl@{}} n_{1} = \frac{s(s+1)}{2}, \quad n_{2}& =& (\ell-s-1)(\ell -s-2), \quad d = n-n_{1}-n_{2},\\ n-2d - n_{1} &=& 5 s^{2} - s(4\ell -7) + \ell^{2}-5\ell+4 . \end{array} $$

If n − 2d − n₁ ≥ 0 then

$$ s \leq \frac{1}{10}\left( 8\ell -13 - \sqrt{24\ell^{2}- 8\ell+9}\right) \quad \text{or} \quad s \geq \frac{1}{10}\left( 8\ell -13 + \sqrt{24\ell^{2}-8\ell+9}\right) . $$

As s ≥ 1, the first inequality is possible only if ℓ ≥ 8. The second inequality is impossible since its right hand side is bigger than ℓ − 4 and s is at most ℓ − 4.

By the above equalities,

$$ n -2d - n_{2} = 2s^{2} + (-2\ell + 4)s - 2\ell+2. $$

If the left hand side is nonnegative then

$$ s \leq -1 \quad \text{or} \quad s\geq \ell-1 . $$

These inequalities are impossible since s is positive and smaller than ℓ − 3, whence Assertion (iii) of Proposition A.1.

1.4 A.4 The Exceptional Cases

Set ${\mathfrak {l}}:={\mathfrak {l}}_{X}$, ${\mathfrak {d}} := {\mathfrak {d}}_{X}$. Then $2d = \dim \hskip .125em \mathfrak {g}-\dim \hskip .125em {\mathfrak {l}}$. For each case, we give all the possible dimensions of ${\mathfrak {l}}$ when |X| = ℓ − 1.

• (a) The algebra $\mathfrak {g}$ has type G₂. Then X is connected, whence Assertion (iv) of Proposition A.1 for this case.

• (b) The algebra $\mathfrak {g}$ has type F₄. In this case n = 24 and

  $$ \dim\hskip .125em {\mathfrak{l}} \in \{12,22\} \quad \text{whence} \quad 2d \in \{40,30\}$$

  and Assertion (iv) of Proposition A.1 for this case.

• (c) The algebra $\mathfrak {g}$ has type E₆. In this case n = 36 and

  $$ \dim\hskip .125em {\mathfrak{l}} \in \{20,28,36,46\} \quad \text{whence} \quad 2d \in \{58,50,42,32 \}$$

  and Assertion (iv) of Proposition A.1 for this case since ${\mathfrak {d}}$ is simple of type D₅ when 2d = 32.

• (d) The algebra $\mathfrak {g}$ has type E₇. In this case n = 63 and

  $$ \dim\hskip .125em {\mathfrak{l}} \in \{27,33,39,49,67,79\} \quad \text{whence} \quad 2d \in \{106,100,94,84,66,54\}$$

  and Assertion (iv) of Proposition A.1 for this case since ${\mathfrak {d}}$ is simple of type E₆ when 2d = 54.

• (e) The algebra $\mathfrak {g}$ has type E₈. In this case n = 120 and

  $$\dim {\mathfrak{l}} \in \{36,40,52,54,64,82,92,134\} \text{ whence } 2d \in \{212,208,196,194,184,166,156,114\}$$

  and Assertion (iv) of Proposition A.1 for this case since $\mathfrak {d}$ is simple of type E₇ when 2d = 114.