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A differential algebra is an algebra (a vector space with a vector-valued product defined between any two vectors) equipped with a derivative (a function from vectors to vectors that is linear and satisfies the Leibniz product rule with respect to the product of the algebra). In the exterior algebra of differential forms on a manifold, the vectors are ...

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The set of differential forms along with addition, scalar multiplication, and the exterior product is an algebra (over a field). Add the exterior derivative and you get a differential algebra. That is, an algebra with an operation which is linear and obeys Leibniz's rule.

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In my opinion, the best way to see that $\det(A^T)=\det(A)$ is via the decomposition of $A$ into shears, transpositions and diagonal matrices. Every square matrix $A$ (regardless of the field) can be decomposed into a product of the form $A=E_sE_{s-1}\cdots E_1DF_1F_2\cdots F_t$, where $D$ is a diagonal matrix, each $E_i$ is an elementary matrix ...

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If $A$ is diagonalizable, so that there exists an invertible matrix $C$ such that $D=CAC^{-1}$ is diagonal, then $D=D^t=(CAC^{-1})^t=C^{-t}A^tC^t$. Since $D$ and $D^t$ have the same determinant, simply because the two matrices are in fact equal, it follows at once from this that $A$ and $A$ and $A^t$ have the same determinant. As diagonalizable matrices are ...

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The determinant of a matrix does not change when you compute it via cofactor expansion along column or row. Thus expanding along a row in $A$ is equivalent to expanding along a column in $A^t$. I'm not sure if this is what you meant by "using invertibility".

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We do, but for historical reasons they are called (linear) isometries.

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I'll propose to you another (slightly different, but isomorphic) definition of the adjugate (classical adjoint). Im borrowing from section 8 of http://people.reed.edu/~jerry/332/27exterior.pdf . Let $f:V\rightarrow V$ (with $n$ the dimension of $V$). We have a canonical isomorphism $\phi:V=\wedge^1 V\rightarrow\mathrm{Hom}(\wedge^{n-1} V,\wedge^n V)$ ...

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Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why $\det A=\det(E_1)\cdots\det(E_k)\det(D)\det(F_1)\cdots\det(F_\ell)$ and the ...

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To prove that the two (generalized) volumes are equal is going to require some algebra -- I don't see a way around that (because the transpose is defined algebraically). This proof doesn't require the use of matrices or characteristic equations or anything, though. I just use a geometric definition of the determinant and then an algebraic formula relating ...

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you can read S.S.Chern's"lectures on differential geometry". He did very well on that book

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A very nice gentle (albeit abstract) introduction to forms and connections can be found in R.W.R Darling's Differential Forms and Connections (1), a more physics based text book would have to be Nakahara's Geometry, Topology & Physics (2) - these helped me greatly when I had a similar need to you. Good Luck! (1) ...

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They both may be right. If wedge product differs (Similar problem) and we set the definition of exterior derivative as $$d\omega=\sum_{I}d\omega_I\wedge dx^I,$$ then $d$ may differs as well (cause $\wedge$ appears). If we take axiomatic approach to exterior derivative, then one of axioms says ...

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The second product you give is equivalent to a tensor product; if it were an exterior product instead, the resulting matrix would be antisymmetric. Also notice that the trace of that matrix is the inner product. So the inner product can be identified with the trace, and the exterior product could be identified with the antisymmetric part of this matrix, ...

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Your derivation is right and equation [1] cannot be correct without further assumptions. In fact the exterior derivative of a 1-form can be shown to be $$\mathrm{d}\omega(X,Y)= X(\omega(Y)) - Y(\omega(X))-\omega([X,Y])$$ which is equivalent to what you wrote.

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