Bourbaki

Corollary

Tag 080Q, permanent. Cite it as [Bourbaki, Tag 080Q].

Topological Vector Spaces, V, p. 34, § 3. Tensor product of hilbertian spaces

Suppose that the hilbertian space E is the direct sum of two orthogonal subspaces M and N. The canonical isomorphism g from Λ(M)⊗Λ(N)\Lambda(M) \otimes \Lambda(N) onto Λ(E)\Lambda(E) (A, III, § 7, No. 7) extends in a unique way to a hilbertian space isomorphism from Λ^(M)⊗^2Λ^(N)\hat{\Lambda}(M) \hat{\otimes}_2 \hat{\Lambda}(N) onto Λ^(E)\hat{\Lambda}(E).

The proof is analogous to that of the corollary of prop. 4 (V, p. 31).

Let E and F be two hilbertian spaces and u∈L(E;F)u \in \mathscr{L}(E; F). We shall show, as in the case of symmetric powers S^n(E)\hat{S}^n(E) (V, p. 32) that the linear mapping Λ^n(u)\hat{\Lambda}^n(u) from Λ^n(E)\hat{\Lambda}^n(E) into Λ^n(F)\hat{\Lambda}^n(F) (A, III, § 7, No. 4) extends to a continuous linear mapping Λ^n(u)\hat{\Lambda}^n(u) from Λ^n(E)\hat{\Lambda}^n(E) into Λ^n(F)\hat{\Lambda}^n(F). We have the relations

Λ^n(1E)=1Λ^n(E),(27) \hat{\Lambda}^n(1_E) = 1_{\hat{\Lambda}^n(E)} , \tag{27}
Λ^n(v∘u)=Λ^n(v)∘Λ^n(u)if v belongs to L(F;G),(28) \hat{\Lambda}^n(v \circ u) = \hat{\Lambda}^n(v) \circ \hat{\Lambda}^n(u) \quad \text{if } v \text{ belongs to } \mathscr{L}(F; G) , \tag{28}
∥Λ^n(u)∥≤∥u∥n.(29) \|\hat{\Lambda}^n(u)\| \leq \|u\|^n . \tag{29}

In general we do not have equality in formula (29) (TS, IV, § 6). Finally, we have an isomorphism ψn=ψn,E\psi_n = \psi_{n,E} from Λ^n(E)\hat{\Lambda}^n(E) onto the subspace A‾n′(E)\overline{\mathbf{A}}'_n(E) of T^n(E)\hat{T}^n(E) defined by

ψn(x1∧...∧xn)=1(n!)1/2∑σ∈Snεσxσ(1)⊗...⊗xσ(n).(30) \psi_n(x_1 \wedge ... \wedge x_n) = \frac{1}{(n!)^{1/2}} \sum_{\sigma \in \mathfrak{S}_n} \varepsilon_\sigma x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)} . \tag{30}

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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