Bourbaki

Example

Tag 0A9D, permanent. Cite it as [Bourbaki, Tag 0A9D].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let E be a complete normable space considered as a Lie group. Then the vector space U(E)U(E) is canonically identified with the vector space TS(E) (Differentiable and Analytic Manifolds, R, 13.2.4). Let m:E×E→Em : E \times E \to E be addition on E. Then

m∗:TS(E×E)→TS(E) m_* : TS(E \times E) \to TS(E)
is equal to TS(m) (Differentiable and Analytic Manifolds, R, 13.2.4). For t,t′t, t' in U(E)=TS(E)U(E) = TS(E), the image t∗t′t * t' of the symmetric tensor product t⊗t′t \otimes t' under m∗m_* is therefore TS(m)(t⊗t′t \otimes t'). By Algebra, Chapter IV, § 5, no. 6, Proposition 7, this image is just the product tt′tt' in the algebra TS(E). Thus the algebra U(E)U(E) is identified with the algebra TS(E).

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

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Cites

Transcribed from the printed volume. Not translated.