Library / Lie Groups and Lie Algebras / Chapter VIII / § 9
Example 2
Tag 0160, permanent. Cite it as [Bourbaki, Tag 0160].
Lie Groups and Lie Algebras, chapter VIII, § 9. The formula of Hermann Weyl
Take g \mathfrak{g} g G 2 G_2 G 2 h R ∗ \mathfrak{h}^*_{\mathbf{R}} h R ∗ ( ⋅ ∣ ⋅ ) (\cdot | \cdot ) ( ⋅ ∣ ⋅ ) ( α 1 ∣ α 1 ) = 1 (\alpha_1|\alpha_1) = 1 ( α 1 ∣ α 1 ) = 1 ρ = ϖ 1 + ϖ 2 \rho =\varpi_1+\varpi_2 ρ = ϖ 1 + ϖ 2
( ϖ 1 ∣ α 1 ) = 1 2 , ( ϖ 1 ∣ α 2 ) = 0 , ( ϖ 1 ∣ α 2 + α 1 ) = 1 2
(\varpi_1|\alpha_1) =\frac{1}{2},(\varpi_1|\alpha_2) = 0,(\varpi_1|\alpha_2+\alpha_1) =\frac{1}{2}
( ϖ 1 ∣ α 1 ) = 2 1 , ( ϖ 1 ∣ α 2 ) = 0 , ( ϖ 1 ∣ α 2 + α 1 ) = 2 1 ( ϖ 1 ∣ α 2 + 2 α 1 ) = 1 , ( ϖ 1 ∣ α 2 + 3 α 1 ) = 3 2 , ( ϖ 1 ∣ 2 α 2 + 3 α 1 ) = 3 2
(\varpi_1|\alpha_2+ 2\alpha_1) = 1,(\varpi_1|\alpha_2+ 3\alpha_1) =\frac{3}{2},(\varpi_1|2\alpha_2+ 3\alpha_1) =\frac{3}{2}
( ϖ 1 ∣ α 2 + 2 α 1 ) = 1 , ( ϖ 1 ∣ α 2 + 3 α 1 ) = 2 3 , ( ϖ 1 ∣2 α 2 + 3 α 1 ) = 2 3 ( ϖ 2 ∣ α 1 ) = 0 , ( ϖ 2 ∣ α 2 ) = 3 2 , ( ϖ 2 ∣ α 2 + α 1 ) = 3 2
(\varpi_2|\alpha_1) = 0,(\varpi_2|\alpha_2) =\frac{3}{2},(\varpi_2|\alpha_2+\alpha_1) =\frac{3}{2}
( ϖ 2 ∣ α 1 ) = 0 , ( ϖ 2 ∣ α 2 ) = 2 3 , ( ϖ 2 ∣ α 2 + α 1 ) = 2 3 ( ϖ 2 ∣ α 2 + 2 α 1 ) = 3 2 , ( ϖ 2 ∣ α 2 + 3 α 1 ) = 3 2 , ( ϖ 2 ∣ 2 α 2 + 3 α 1 ) = 3
(\varpi_2|\alpha_2+ 2\alpha_1) =\frac{3}{2},(\varpi_2|\alpha_2+ 3\alpha_1) =\frac{3}{2},(\varpi_2|2\alpha_2+ 3\alpha_1) = 3
( ϖ 2 ∣ α 2 + 2 α 1 ) = 2 3 , ( ϖ 2 ∣ α 2 + 3 α 1 ) = 2 3 , ( ϖ 2 ∣2 α 2 + 3 α 1 ) = 3 Thus, if n 1 , n 2 n_1, n_2 n 1 , n 2 ≥ 0 \geq 0 ≥ 0 n 1 ϖ 1 + n 2 ϖ 2 n_1\varpi_1+n_2\varpi_2 n 1 ϖ 1 + n 2 ϖ 2
( 1 + n 1 / 2 1 2 ‾ ) ( ( 1 + 3 n 2 / 2 3 2 ‾ ) ( ( 1 + n 1 / 2 + 3 n 2 / 2 1 2 ‾ + 3 2 ) ( ( 1 + n 1 + 3 n 2 / 2 1 + 3 2 ‾ )
(1 +\frac{n_1/2}{1\overline{2}})((1 +\frac{3n_2/2}{3\overline{2}})((1 +\frac{n_1/2 + 3n_2/2}{_{1\overline{2}} + \frac{3}{2}})((1 +\frac{n_1 + 3n_2/2}{1 +^{3\overline{2}}})
( 1 + 1 2 n 1 /2 ) (( 1 + 3 2 3 n 2 /2 ) (( 1 + 1 2 + 2 3 n 1 /2 + 3 n 2 /2 ) (( 1 + 1 + 3 2 n 1 + 3 n 2 /2 ) × ( 1 + 3 n 1 / 2 + 3 n 2 / 2 3 2 ‾ + 3 2 ) ( ( 1 + 3 n 1 / 3 2 2 + 3 + 3 n 2 )
\times (1 +\frac{3n_1/2 + 3n_2/2}{_{3\overline{2}} + \frac{3}{2}})((1 +3n_1\frac{/_3}{2}2 + 3+ 3n_2)
× ( 1 + 3 2 + 2 3 3 n 1 /2 + 3 n 2 /2 ) (( 1 + 3 n 1 2 / 3 2 + 3 + 3 n 2 ) = ( 1 + n 1 ) ( 1 + n 2 ) ( 1 + n 1 + 3 n 2 4 ) ( ( 1 + 2 n 1 + 3 n 2 5 ) ( ( 1 + n 1 + n 2 2 )
= (1 +n_1)(1 +n_2)(1 +\frac{n_1 + 3n_2}{4})((1 +\frac{2n_1 + 3n_2}{5})((1 +\frac{n_1 + n_2}{2})
= ( 1 + n 1 ) ( 1 + n 2 ) ( 1 + 4 n 1 + 3 n 2 ) (( 1 + 5 2 n 1 + 3 n 2 ) (( 1 + 2 n 1 + n 2 ) × ( 1 + n 1 + 2 n 2 3 )
\times (1 +\frac{n_1 + 2n_2}{3})
× ( 1 + 3 n 1 + 2 n 2 ) ( 1 + n 1 ) ( 1 + n 2 ) ( 2 + n 1 + n 2 ) ( 3 + n 1 + 2 n 2 ) ( 4 + n 1 + 3 n 2 ) ( 5 + 2 n 1 + 3 n 2 )
(1+n_1)(1+n_2)(2+n_1+n_2)(3+n_1+2n_2)(4+n_1+3n_2)(5+2n_1+3n_2)
( 1 + n 1 ) ( 1 + n 2 ) ( 2 + n 1 + n 2 ) ( 3 + n 1 + 2 n 2 ) ( 4 + n 1 + 3 n 2 ) ( 5 + 2 n 1 + 3 n 2 ) =.
5!
In particular, the fundamental representation of highest weight ϖ 1 \varpi_1 ϖ 1 ϖ 2 \varpi_2 ϖ 2
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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Transcribed from the printed volume. Not translated.