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With T: R^m -->R^n be linear transformation T(x) = B*x and if psi sub I is an elementary alternating k-tensor on R^n, then T*psisub I has the form:

$$ T^**\psi_I $$ = sigma sub [J] cJ*psi[J] where psiJ are the elementary alternating k-tensors on R^m and

where $I = (i_1,\ldots, i_k)$ and we can let $I_\sigma = (i_{\sigma(1)},\ldots, i_{\sigma(k)})$.

I'm trying to determine the coefficients of C sub J.

My proof:

T*f(x) = f(T(x)) = f(Bx) = ABx so the matrix T*f is AB

And if f = sigma sub [I] dsubI * psisub I is an alternating k-tensor on R^n, how can T*f be expressed in terms of the elementary k-tensors on R^m?

Thanks

  • 1
    Why don't you write in TeX?2012-03-20
  • 0
    I'm trying to learn2012-03-20
  • 2
    What is $\phi$? What is $\psi$?2012-03-20
  • 0
    The tensors phi sub I are elementary k-tensors on V corresponding to a basis a1,...an for V.2012-03-20
  • 0
    Psi sub I = summation sub sigma (sgn sigma) * (phi sub I)^sigma where the summation extends over all sigma as an element of S sub k2012-03-20
  • 0
    Then of course the two are equal. You have defined that $\psi_I = \sum_{\sigma\in S_k}\mbox{sgn} \sigma\phi_{I_\sigma}$.2012-03-20
  • 0
    Since the question has "differential forms" in the title, are you trying to figure out the details of the alternating map $A$ from $T^0_k$ to the alternating $k$-forms, $A\psi(v_1,\ldots,v_k)=\sum_{\sigma\in S_k}\mbox{sgn}\sigma\ \psi(v_{\sigma(1)},\ldots,v_{\sigma(k)})$?2012-03-20
  • 2
    Mary, after visiting [this Latex site](http://www.codecogs.com/latex/eqneditor.php), you should be able to create equations without having to write "psi sub I" etc.2012-03-20

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