Phys 509 - Problem Set 3
نویسنده
چکیده
1 Innitesimal Homotopy The innitesimal homotopy relation reads: LX ω = d (iX ω) + iX (dω) Using this, we wish to evaluate LX (dω) and d (LX ω). First: LX dω = d (iX dω) + iX ddω However, since ddω = d 2 ω = 0, this leaves us with: LX dω = d (iX dω) On the other hand, we have: d (LX ω) = d (d (iX ω) + iX dω) = d 2 (iX ω) + d (iX dω) Once again, d 2 (iX ω) = 0; this leave us with: d (LX ω) = d (iX dω) Comparing the two yields the desired equality:
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Phys 509 - Problem Set 6
We show that CG ({g}) is a subgroup CG ({g}) ≤ G. • Closure: For x, y ∈ GG ({g}), then xg = gx and yg = gy, so in particular gxy = xgy = xyg, so xy ∈ CG ({g}) • Inverses: For x ∈ CG ({g}) ⊆ G, ∃x−1 ∈ G s.t. xx−1 = x−1x = e, since G is a group. For g−1 ∈ G exists, then we can write gx−1g−1x = g (gx) x = g (xg) x = gg−1x−1x = e, which shows gx−1 = x−1g, so x−1 ∈ CG ({g}) • Identity: For e ∈ G, eg...
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