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# 02 - Lecture 02 Suppose k is an ideal of g The...

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Lecture 02 Suppose k is an ideal of g . The corresponding quotient algebra is constructed as follows: Let g / k be the quotient vector space. Define a bracket [ · , · ] on g / k by [ x + k , y + k ] := [ x, y ] + k . This bracket is well-defined. Indeed, suppose x 0 + k = x + k and y 0 + k = y + k . Then [ x 0 , y 0 ] - [ x, y ] = [ x 0 - x, y 0 ] + [ x, y 0 - y ] . Since k is an ideal, each of [ x 0 - x, y 0 ] and [ x, y 0 - y ] is in k . Since k is a subspace, [ x 0 - x, y 0 ] + [ x, y 0 - y ] k . It is an easy exercise to verify that [ · , · ] defined as above on g / k is bilinear, alternating and satisfies the Jacobi identity. Example 1. The quotient of g by its derived algebra [ g , g ] is abelian (this is obvious from the definition of derived algebra). In fact, [ g , g ] is the smallest ideal k for which g / k is abelian: If g / k is abelian, then for each x, y g , [ x, y ] + k = [ x + k , y + k ] = 0 so that [ x, y ] k . By considering the linear span of all x, y g , we see that [ g , g ] k . Example 2. Z ( g /Z ( g )) need not be zero. Example? See the Heisenberg algebra (next lecture). Definition 1. A homomorphism from g to h is a linear map φ : g h that preserves brackets: x, y g φ ([ x, y ]) = [ φ ( x ) , φ ( y )] . An isomorphism from g to h is a homomorphism φ : g h for which there is a homomor- phism ψ : h g such that both ψ φ = id g and φ ψ = id h (alternatively, φ is a bijective homomorphism).

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02 - Lecture 02 Suppose k is an ideal of g The...

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