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In page 2 of Hans Samelson's Notes on Lie Algebras, the text gives an example of Lie algebra $A_L$.

Let $A$ be an algebra over $\mathbb{F}$ (a vector space with an associative multiplication $X\cdot Y$). We make $A$ into a Lie algebra $A_L$ (also called $A$ as Lie algebra) by defining $[X,Y]=X\cdot Y-Y\cdot X$. The Jacobi identity holds; just "multiply out".

Multiplying out the Jacobi identity and using associativity of $\cdot$, we have

$X\cdot Y\cdot Z-X\cdot Z\cdot Y+Y\cdot Z\cdot X-Y\cdot X\cdot Z+Z\cdot X\cdot Y-Z\cdot Y\cdot X=0$.

Assuming only associativity, I can't see why this holds.

Conifold
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PJ Miller
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1 Answers1

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The Jacobi identity will give the following:

$$[[x,y],z]+[[y,z],x]+[[z,x],y]=\\(xy-yx)z-z(xy-yx)+(yz-zy)x-x(yz-zy)+(zx-xz)y-y(zx-xz)$$

See if you can multiply now, and use associativity to show that the above expression vanishes.

Jared
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  • Oh whoops, in $[X[YZ]]$ I just expanded Jacobi on the inner one, but thought the outer one is just multiplication. Thanks! – PJ Miller Aug 04 '13 at 02:59