The Gram inner product on ${\mathrm{\Lambda }}^{k}V$

Let $V$ be a finite-dimensional real inner product space (Real and complex inner product spaces, with the inner product linear in the first argument) and $k\ge 0$. On decomposable wedges of The $k$th exterior power as the tensor-power quotient by repeated-vector relations put

$⟨v_1\wedge \cdots \wedge v_k,w_1\wedge \cdots \wedge w_k⟩:=\det (⟨v_i,w_j⟩{)}_{i,j\le k},$

the determinant (For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix) of the $k\times k$ matrix of pairings. For $k=0$ both sides are the empty determinant, which is $1$.

For each fixed list $(w_1,\dots ,w_k)$, the assignment $(v_1,\dots ,v_k)\mapsto \det (⟨v_i,w_j⟩)$ is $k$-linear (each entry is linear in the corresponding $v_i$, and the determinant is multilinear in its rows) and alternating (if $v_i=v_{i^{\mathrm{\prime }}}$, two rows of the matrix are equal, so the determinant vanishes). It therefore descends to a linear functional on ${\mathrm{\Lambda }}^{k}V$ in the first slot, and symmetrically in the second slot. The resulting bilinear pairing is the Gram inner product on ${\mathrm{\Lambda }}^{k}V$; that it is an inner product in the sense of Real and complex inner product spaces, with the inner product linear in the first argument is proved in The Gram formula gives a well-defined positive-definite inner product on exterior powers, and $\mathrm{\parallel }{v}_1\wedge \cdots \wedge v_k{\mathrm{\parallel }}^2$ is the Gram determinant ↗, which also discharges the well-definedness obligation recorded above.