By Satya Deo

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**Sample text**

Ka ... i p−1 ω jb ib ∧ ωka ia b=a = 12 R ja ia lk ai1 ... ja ... i p−1 ωk ∧ ωl . 3 Bochner–Weitzenb¨ock formulas 23 We now claim that the second term on the left-hand side is identically 0. Indeed, since ai1 ... jb ... ka ... i p−1 ω jb ib ∧ ωka ia b=a = ai1 ... jb ... ka ... i p−1 ω jb ib ∧ ωka ia + ba the claim follows by interchanging the roles of ka and jb in the second term. Hence ai1 ... i p−1 , jk ωk ∧ ω j = 12 R ja ia lk ai1 ... ja ...

Suppose x0 ∈ M is the point where P achieves its maximum. If |∇u(x 0 )| = 0 we may rotate the frame so that u 1 (x0 ) = |∇u(x0 )|. Differentiating in the ei direction yields 1 2 Pi = u j u ji + cuu i . 5 Poincar´e inequality and the first eigenvalue 43 At the point x0 , this yields 0 = u 1 (u 11 + cu) and u ji u ji ≥ u 211 = c2 u 2 . 2), the definition of P, and evaluating at x 0 , we have 0≥ 1 2 P = u ji u ji + u j u jii + cu i2 + cu u ≥ c2 u 2 + u j ( u) j + R ji u j u i + cu i2 − cλu(u + a) ≥ c2 u 2 − λu i2 + cu i2 − cλu(u + a) = (c − λ) u i2 + cu 2 − acλu ≥ aλP(x0 ) − acλ.

Im− p )) ai1 ... kθ ... i p αth slot ωiα kθ . Using the skew symmetry ωiα kθ = − ωkθ iα , we conclude that dβ = sgn(σ (I, I c )) ai1 ... i α ... i p ,i α ωiα ∧ ωim ∧ . . ∧ ωi p+1 , hence ∗dβ = sgn(σ (I, I c )) sgn(σ (i α , i m , . . , i p+1 , i p , . . , iˆα , . . , i 1 )) × ai1 ... iα ... i p ,iα ωi p ∧ . . ∧ ωˆ iα ∧ . . ∧ ωi1 = (−1)m−α+ p(m− p) ai1 ... i p ,iα ωi p ∧ . . ∧ ωˆ iα ∧ . . ∧ ωi1 , and (−1)α+ p δω = 2 +1 ai1 ... iα ... i p ,i α ωi p ∧ . . ∧ ωˆ iα ∧ . . ∧ ωi1 . 1≤α≤ p Computing directly gives − ω = δ(ai1 ...