Choose the chain basis $v_0,\ldots,v_6$ from the preceding part, normalized by $v_1=f_1v_0$, $v_2=f_2v_1$, $v_3=f_1v_2$. The associated <sl2 Lie algebra> strings give $e_1v_1=v_0$, $e_1v_3=2v_2$ and $e_2v_2=v_1$; other raising actions used below vanish because their proposed weights do not occur.

In the <exterior square>, $v_0\wedge v_1$ is a <highest-weight vector> of weight $2\omega_1-\alpha_1=\omega_2$: $e_1$ produces $v_0\wedge v_0=0$, and $e_2$ kills both factors. Also
$$
w=v_0\wedge v_3-2v_1\wedge v_2
$$
is a nonzero <highest-weight vector> of weight $\omega_1$. Its two $e_1$ terms cancel, and its $e_2$ image is zero. The <Weyl complete reducibility theorem> supplies irreducible summands with those weights; their dimensions $14+7=21$ exhaust $\dim\Lambda^2V$. Thus
$$
\boxed{\Lambda^2V=L(\omega_2)\oplus L(\omega_1).}
$$
For the <symmetric square>, $v_0\otimes v_0$ is a <highest-weight vector> of weight $2\omega_1$, giving a 27-dimensional summand. The <longest Weyl-group element> of $G_2$ acts as minus the identity, so the <highest weight of a dual representation> is $-w_0\omega_1=\omega_1$: $V$ is self-dual. A nonzero intertwiner $V\to V^*$ gives a <nondegenerate> <Lie-invariant bilinear form>. It is symmetric, since an <alternating bilinear form> cannot be nondegenerate in odd <dimension>. Its inverse is a nonzero invariant vector in $S^2V$, supplying the scalar summand. Since $27+1=28=\dim S^2V$,
$$
\boxed{S^2V=L(2\omega_1)\oplus L(0).}
$$
These are the <exterior and symmetric squares of the seven-dimensional G2 representation>; all highest-weight and dimension claims are specified rather than relying on an unstated cross-product identity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 102 / 5 / iii / Solution

For $V=L(\omega_1)$ with the <G2 root system> numbered short-first,
$$
\Lambda^2V=L(\omega_2)\oplus L(\omega_1),\qquad S^2V=L(2\omega_1)\oplus L(0).
$$
In the chain basis of the <crystal of the seven-dimensional G2 representation>, normalize $v_1=f_1v_0$, $v_2=f_2v_1$, $v_3=f_1v_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero <highest-weight vectors> of weights $\omega_2$ and $\omega_1$; the identities $e_1v_1=v_0$, $e_1v_3=2v_2$, $e_2v_2=v_1$ verify both raising conditions. The <Weyl complete reducibility theorem> and <G2 dimension polynomial> exhaust the twenty-one dimensions of the <exterior square>. In the <symmetric square>, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a <Lie-invariant bilinear form>, which is symmetric because a <nondegenerate> <alternating bilinear form> cannot have odd <dimension>; its inverse gives the remaining invariant line.

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 Incoming links: Exterior and symmetric squares of the seven-dimensional G2 representation