Notes for AKT-140307/0:41:01: Difference between revisions

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'''Proposition: '''$CS(A^g) = CS(A)$
'''Proposition: '''$CS(A^g) = CS(A)+\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$
$$CS(A^g)=\int_\mathbb{R^3} Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g)$$
$$CS(A^g)=\int_\mathbb{R^3} Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g)$$
$$Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g) =$$
$$Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g) =$$
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$$g^{-1} A g \wedge d g^{-1} \wedge d g = g^{-1} A g \wedge d g^{-1} g \wedge g^{-1} d g = - g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g $$
$$g^{-1} A g \wedge d g^{-1} \wedge d g = g^{-1} A g \wedge d g^{-1} g \wedge g^{-1} d g = - g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g $$


Combining this with the fact that the trace is invariant under cyclic permutations show that the
Combining this with the fact that the trace is invariant under cyclic permutations show that


$$Tr(g^{-1} A \wedge (d A) g - 2 g^{-1} A \wedge A \wedge d g - 2 g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g +g^{-1} d g \wedge g^{-1} d A g - g^{-1} d g \wedge g^{-1} A \wedge d g + g^{-1} d g \wedge d g^{-1} \wedge d g) +$$
$$Tr(g^{-1} A \wedge (d A) g - 2 g^{-1} A \wedge A \wedge d g - 2 g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g +g^{-1} d g \wedge g^{-1} d A g - g^{-1} d g \wedge g^{-1} A \wedge d g + g^{-1} d g \wedge d g^{-1} \wedge d g) +$$
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$$Tr(g^{-1} d g \wedge d(g^{-1} A) g) = Tr(d g \wedge d(g^{-1} A)) = Tr(d (gd(g^{-1} A))) = d Tr(g d(g^{-1} A))$$
$$Tr(g^{-1} d g \wedge d(g^{-1} A) g) = Tr(d g \wedge d(g^{-1} A)) = Tr(d (gd(g^{-1} A))) = d Tr(g d(g^{-1} A))$$
This shows that
This shows that
$$CS(A^g)=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + g^{-1} A \wedge A \wedge A g + \frac13 g^{-1} d g \wedge d g^{-1} \wedge d g) +\int_\mathbb{R^3} d Tr(g d(g^{-1} A))$$
$$CS(A^g)=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g + \frac13 g^{-1} d g \wedge d g^{-1} \wedge d g) +\int_\mathbb{R^3} d Tr(g d(g^{-1} A))$$

A similar argument shows that
If we assuming that A and g are compactly supported then by Stokes' theorem
⚫
$$Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g) = -Tr(g^{-1} d g \wedge d g^{-1} \wedge d g) = dTr( $$
$$\int_\mathbb{R^3}d Tr(g d(g^{-1} A)) = 0$$
to be completed

So

⚫
$$CS(A^g)=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g + \frac13 g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$
$$=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g) +\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$
$$=\int_\mathbb{R^3}Tr(A \wedge d A + \frac23 A \wedge A \wedge A) +\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$
$$=CS(A)+\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$

If we define $CS_1(A,g)=CS(A^g)-CS(0^g)$
Then
$$CS_1(A,g) = CS(A^g) - \frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g) = CS_1(A,I) = CS(A)$$

Latest revision as of 19:15, 24 August 2018

Proposition: $CS(A^g) = CS(A)+\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$ $$CS(A^g)=\int_\mathbb{R^3} Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g)$$ $$Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g) =$$ $$Tr((g^{-1} A g + g^{-1} dg)\wedge d(g^{-1} A g + g^{-1} dg)+\frac23((g^{-1} A g + g^{-1} dg)\wedge(g^{-1} A g + g^{-1} dg)\wedge(g^{-1} A g + g^{-1} dg)))=$$ $$Tr(g^{-1} A \wedge (d A) g + g^{-1} A g \wedge d g^{-1} \wedge g^{-1} A g + g^{-1} d g \wedge g^{-1} (d A) g - g^{-1} d g \wedge g^{-1} A \wedge d g +$$ $$g^{-1} d g \wedge d g^{-1} \wedge A g - g^{-1} A \wedge A \wedge d g + g^{-1} A g \wedge d g^{-1} \wedge d g + g^{-1} d g \wedge d g^{-1} \wedge d g) +$$ $$\frac23 Tr( g^{-1} A \wedge A \wedge A g + g^{-1} A \wedge A \wedge d g + g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} A g + g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g +$$ $$g^{-1} d g \wedge g^{-1} A \wedge A g + g^{-1} d g \wedge g^{-1} A g \wedge g^{-1} d g + g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} A g + g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g) $$

Now $0 = d (g^{-1} g) = (d g) g^{-1} + g d g^{-1}$

So $(dg) g^{-1} = - g d g^{-1}$

Applying this to the fifth and seventh terms of the equation above yields $$ g^{-1} d g \wedge d g^{-1} \wedge A g = g^{-1} d g \wedge d g^{-1} g \wedge g^{-1} A g = - g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} A g$$ and $$g^{-1} A g \wedge d g^{-1} \wedge d g = g^{-1} A g \wedge d g^{-1} g \wedge g^{-1} d g = - g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g $$

Combining this with the fact that the trace is invariant under cyclic permutations show that

$$Tr(g^{-1} A \wedge (d A) g - 2 g^{-1} A \wedge A \wedge d g - 2 g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g +g^{-1} d g \wedge g^{-1} d A g - g^{-1} d g \wedge g^{-1} A \wedge d g + g^{-1} d g \wedge d g^{-1} \wedge d g) +$$ $$ \frac23 Tr( g^{-1} A \wedge A \wedge A g + 3 g^{-1} A \wedge A \wedge d g + 3 g^{-1} A g \wedge g^{-1} d g \wedge g^{-1} d g +g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g) =$$ $$Tr(g^{-1} A \wedge d A g + g^{-1} d g \wedge g^{-1} d A g - g^{-1} d g \wedge g^{-1} A \wedge d g+ g^{-1} d g \wedge d g^{-1} \wedge d g) + \frac23 Tr( g^{-1} A \wedge A \wedge A g + g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$ Now deal with the extra terms $$Tr(g^{-1} d g \wedge g^{-1} (d A) g ) = Tr(g^{-1} d g \wedge d(g^{-1} A) g - g^{-1} d g \wedge d g^{-1} \wedge A g) = Tr(g^{-1} d g \wedge d(g^{-1} A) g + g^{-1} d g \wedge g^{-1} A \wedge d g)$$ Finally $$Tr(g^{-1} d g \wedge d(g^{-1} A) g) = Tr(d g \wedge d(g^{-1} A)) = Tr(d (gd(g^{-1} A))) = d Tr(g d(g^{-1} A))$$ This shows that $$CS(A^g)=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g + \frac13 g^{-1} d g \wedge d g^{-1} \wedge d g) +\int_\mathbb{R^3} d Tr(g d(g^{-1} A))$$

If we assuming that A and g are compactly supported then by Stokes' theorem $$\int_\mathbb{R^3}d Tr(g d(g^{-1} A)) = 0$$

So

$$CS(A^g)=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g + \frac13 g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$ $$=\int_\mathbb{R^3}Tr(g^{-1} A \wedge (d A) g + \frac23 g^{-1} A \wedge A \wedge A g) +\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$ $$=\int_\mathbb{R^3}Tr(A \wedge d A + \frac23 A \wedge A \wedge A) +\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$ $$=CS(A)+\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$

If we define $CS_1(A,g)=CS(A^g)-CS(0^g)$ Then $$CS_1(A,g) = CS(A^g) - \frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g) = CS_1(A,I) = CS(A)$$

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