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Duke MATH 378 - More on the function F.

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1. More on the function F.LetZ = {(z, A) ∈ R × Sym(Rn) : A is invertible and |z| ||A−1|| < 1}and note that Z is open. LetF(z, A) = trace A ◦ (1 − z A)−1for (z, A) ∈ G.Recall that ifinv : GL(Rn) → GL(Rn)is inversion then∂inv(A)(B) = −A−1◦ B ◦ A−1whenver A ∈ GL(Rn) and B ∈ gl(Rn).We have∂F(z, A)(1, 0) = trace A ◦ (1 − z A)−1◦ A ◦ (1 − z A)−1= trace A2◦ (1 − z A)−2= A2• (1 − z A)−2as well as∂F(z, A)(0, B) = trace B ◦ (1 − z A)−1+ A ◦ (1 − z A)−1◦ (zB) ◦ (1 − z A)−1= trace B ◦ (1 − z A)−1+ B ◦ (1 − z A)−1◦ z A ◦ (1 − z A)−1= trace B ◦ (1 − z A)−1◦³1 + z A ◦ (1 − z A)−1´= trace B ◦ (1 − z A)−2= B • (1 − z A)−2.(1)2. Gory formulae, part one.Suppose v is a smooth function on some open subset Ω of Rn. Letw(x, t) =|∇vt(x)|22for (x, t) ∈ Ω.I claim that(2) ∇wt= ∂(∇vt)(∇vt).Here the right hand side evaluated at x ∈ Ω equals∂(∇vt)(x)((∇vt)(x)).Reasonable, huh? Also,(3) ∂(∇wt) = ∂(∂( ∇vt))(∇vt) + ∂(∇vt) ◦ ∂(∇vt).This means that if x ∈ Ω and h ∈ Rnthen∂(∇wt)(x)(h) = ∂(∂(∇vt)(x))(∇vt(x))(h) +¡∂(∇vt)(x) ◦ ∂(∇vt)(x)¢(h).12Indeed, suppose x ∈ Ω and h ∈ Rn. Then∇wt(x) • h = ∂wt(x)(h)= ∂(∇vt)(x)(h) • ∇vt(x)= ∂(∇vt)(x)(∇vt(x)) • h;this verifies (2). To verify (3) we use (2) to obtain∂(∇wt)(x)(h) = ∂(∂(∇vt)(x)(h)(∇vt(x)) + ∂(∇vt)(x)(∂(∇vt)(x)(h))= ∂(∂(∇vt))(x)((∇vt)(x)(h) + ∂(∇vt)(x)(∂(∇vt)(x)(h)).3. Gory formulae, part two.Suppose Γ is an open subset of Rn× R, v : Γ → R is smooth and˙vt= F(vt, ∂(∇vt)).Let w : Γ → R b e such thatwt=|∇vt|22and letG(x, t) = F(vt(x), ∂(∇vt)(x)) for (x, t) ∈ Γ.Then˙wt= ∇ ˙vt• ∇vt= ∇Gt• ∇vt.For any j ∈ {1, . . . , n} we have∂jGt= ∂F (vt, ∂(∇vt))(∂jvt, ∂j(∂(∇vt)))=³∂jvt(∂(∇vt))2+ ∂j(∂(∇vt))´• Ctwhere C : Γ → Sym(Rn) is such thatCt= (1 − vt∂(∇vt))−2.It follows from (3) and (2) that˙wt=³|∇vt|2(∂(∇vt))2+ ∂(∂(∇vt))(∇vt)´• Ct=³|∇vt|2(∂(∇vt))2+¡∂(∇wt) − ∂(∇vt)2¢´• Ct= Ct• ∂(∇wt) + (∂(∇vt)2• Ct)(2wt− 1).(4)Thus, if c : Γ → R is such thatct= Ct• (∂(∇vt))2and ifz: Γ→Ris such thatzt= |∇vt|2− 1 = 2wt− 1we have˙zt= Ct• ∂(∇zt) +


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