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Advanced TopicsLIE ALGEBRAS(Date Due: 30thJuly)ASSIGNMENT 2Question 1.LetLbe the real vector spaceR3. Givenx,y?L, define[x,y] :=x×y,where×denotes the usualcross productof vectors.Sow thatLis a Lie...

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Advanced TopicsLIE ALGEBRAS(Date Due: 30thJuly)ASSIGNMENT 2Question 1.LetLbe the real vector spaceR3. Givenx,y?L, define[x,y] :=x×y,where×denotes the usualcross productof vectors.Sow thatLis a Lie algebra and determine its structure constants relative to the standardbasis forR3.Question 2.Letdbe a derivation of the Lie algebraL. Show that ifdcommutes with every innerderivation, thend(L)?C(L),whereC(L) denotes thecentreofL.Question 3.Letx?gl(n,F) havendistinct eigenvalues?1,?2,···,?ninF. Prove that the eigenvaluesof adxare then2scalars?i-?j,(1=i, j=n).(Note that onlyn2-n+ 1 scalars are paiwise distinct from each other since?i-?i= 0for alli.)1
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th Advanced Topics LIE ALGEBRAS (Date Due: 30 July) ASSIGNMENT 2 Question 1. 3 Let L be the real vector spaceR . Given x;y2L, de ne [x;y] :=xy; where denotes the usual cross product of vectors. Sow that L is a Lie algebra and determine its structure constants relative to the standard 3 basis forR . Question 2. Let  be a derivation of the Lie algebra L. Show that if  commutes with every inner derivation, then (L)C(L); whereC(L) denotes the centre of L. Question 3. Letx2gl(n;F) haven distinct eigenvalues ; ; ; inF. Prove that the eigenvalues 1 2 n 2 of ad are the n scalars x   ; (1i; jn): i j 2 (Note that only n n + 1 scalars are paiwise distinct from each other since   = 0 i i for all i.) 1

Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021
129 Votes
Exercise-1:
We have
[x, y] = x× y
For being Lie alge
a, this binary operation should satisfy the following condi-
tion:
Bilinear For a, b ∈ R and x, y, z ∈ R3, we have
[ax+ by, z] = (ax+ by)× z = ax× z + by × z
= a(x× z) + b(y × z) = a[x, z] + b[y, z]
1. We have [x, x] = x× x = 0
2. We have
[x, [yz]] + [z[xy]] + [y[zx]]
= x× (y × z) + z × (x× y) + y × (z × z)
= (x.z)y − (x.y)z + (z.y)x− (z.x)y + (y.x)z − (y.z)x
= 0
Hence Jacobi identity satisfies.
This proves that R3 with [x, y] binary operation is Lie alge
a.
Now we have e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) be the basis element.
Now if
[ei, ej ] =
3∑
k=0
ckijek
Then...
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