Back to determining the Lie algebra over the trivial group U(1), as the isomorphism SO(2) as a set of 2 x 2 matrices.

So I can write: $$ M(\theta) = I cos\theta + iXsin\theta $$, where the matrices I and X are as set out in post 119.

Now I expand $$ cos\theta,\; sin\theta $$, and I have:

$$ M(\theta) = I(1 - \frac {\theta^2} {2!} + \frac {\theta^4} {4!} + ... ) + iX(\theta - \frac {\theta^3} {3!} + \frac {\theta^5} {5!} + ... ) $$

$$ = I + i \theta X + \frac {(i\theta X)^2} {2!} + \frac {(i\theta X)^3} {3!} + ... $$,

since $$ X^{2n} = I $$, and $$ X^{2n + 1} = X $$, for n = 1,2,3 ...; and of course powers of

But this is the Taylor expansion of $$ e^{i\theta X} $$! Hence, X is called the generator of the Lie group.

Not quite there yet . . .

So I can write: $$ M(\theta) = I cos\theta + iXsin\theta $$, where the matrices I and X are as set out in post 119.

Now I expand $$ cos\theta,\; sin\theta $$, and I have:

$$ M(\theta) = I(1 - \frac {\theta^2} {2!} + \frac {\theta^4} {4!} + ... ) + iX(\theta - \frac {\theta^3} {3!} + \frac {\theta^5} {5!} + ... ) $$

$$ = I + i \theta X + \frac {(i\theta X)^2} {2!} + \frac {(i\theta X)^3} {3!} + ... $$,

since $$ X^{2n} = I $$, and $$ X^{2n + 1} = X $$, for n = 1,2,3 ...; and of course powers of

*i*alternate: even powers are ±1, odd powers are ±*i*.But this is the Taylor expansion of $$ e^{i\theta X} $$! Hence, X is called the generator of the Lie group.

Not quite there yet . . .

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