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Engel triangularization theorem
Statement
Let be finite-dimensional over any field, and let be nil. Then there is a flag
with and . Equivalently, there is a basis in which every is strictly upper triangular.
Facts & Assumptions
Given: A nil representation of a Lie algebra on a finite-dimensional vector space .
On a nonzero finite-dimensional nil module there is a nonzero vector annihilated by all of (Engel's common-zero-vector lemma).
An invariant subspace and its quotient carry the restricted and induced representations (Subrepresentations, quotient representations, and intertwiners).
The canonical projection onto a vector-space quotient is linear and surjective (The quotient vector space and its canonical projection).
Proof
If , the empty flag has the required property and the empty basis gives the matrix assertion.
Assume and the theorem for all nil modules of dimension smaller than .
By [L1], choose with , and set . This is an invariant one-dimensional subspace annihilated by .
Every induced operator on is nilpotent, since a power of that vanishes on also vanishes on the quotient. By [L2] and step 1.2, the quotient has a flag with . Taking inverse images under the projection [L3] and adjoining gives the required flag of .
Choose vectors adapted to the flag, in the finite sequential sense. The containment says every matrix sends the th basis vector into the span of earlier vectors, hence is strictly upper triangular in this ordering. Conversely a common strictly upper-triangular basis supplies exactly this flag. The zero case was covered in step 1.1.
Depends on
Used by
Dependency tree · two levels
9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Knapp, Lie Groups Beyond an Introduction, Theorem 1.35 (standard reference, not scraped)