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Extra info for Solution manual for Manifolds, tensors, and forms

Example text

Ir j1 ... ir j1 ... ir j1 ... ir j1 ... ir j1 ... ir j1 ... js (+X js ,k )∂i1 ⊗ · · · ⊗ ∂ir ⊗ d x i1 ⊗ · · · ⊗ d x js−1 ⊗ d x k .

Lastly, if U is open in Z then by the continuity of g, g −1 (U ) = π −1 ◦ f −1 (U ) is open in X . So by the definition of the quotient topology, f −1 (U ) is open in Y , so f is continuous. 14 It suffices to show that the Jacobian matrix equals the matrix ( f i j ) representing f itself relative to the standard bases. By linearity, f (x) = f ( x jej) = ei f i j x j , j ij so the i th coordinate function is f i (x) = gives j f i j x j . Taking partial derivatives ∂fi = f i j. 15 Following the hint, we observe that f ◦ f −1 = id implies that (D f )(x) · (D f −1 )(y) = I , which shows that (D f −1 )(y) = (D f (x))−1 .

From the definition of the determinant, n B(e1 ∧ e2 ∧ · · · ∧ en ) = (det B)e1 ∧ e2 ∧ · · · ∧ en . Thus, tr n B = det B, because the trace just sums all the components of the linear operator that map the basis elements to themselves. ) But, from the definition, n B(e1 ∧ e2 ∧ · · · ∧ en ) = Be1 ∧ Be2 ∧ · · · ∧ Ben = (I +z A)e1 ∧ (I +z A)e2 ∧ · · · ∧ (I +z A)en . Now we are finished, because the coefficient of z r consists of a sum of all possible (ordered) wedge products of r Aei ’s and n − r ei ’s.

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