A problem in Elementary Differential Geometry

jdinatale
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My teacher has defined U_1 = \langle1, 0, 0\rangle, U_2 = \langle0, 1, 0\rangle, and U_3 = \langle0, 0, 1\rangle.

So it seems like the function maps L(\langle1, 0, 0\rangle, \langle0, 1, 0\rangle) = a, L(\langle1, 0, 0\rangle, \langle0, 0, 1\rangle) = b,, and L(\langle0, 1, 0\rangle, \langle0, 0, 1\rangle) = c

I'm not sure how that helps me determine what L(\langle v_1, v_2, v_3\rangle, \langle w_1, w_2, w_3\rangle) is.

Could the function perhaps be L(u, v) = L(\langle v_1, v_2, v_3\rangle, \langle w_1, w_2, w_3\rangle) = a\cdot{}v_1 + b\cdot{}w_3 + c\cdot{}v_2

That seems to satisfy our initial conditions.

My textbook doesn't cover this and my teacher hasn't shown an example of how these functions work, so I'm unsure what to do.
 
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jdinatale said:
Joseph.png



My teacher has defined U_1 = \langle1, 0, 0\rangle, U_2 = \langle0, 1, 0\rangle, and U_3 = \langle0, 0, 1\rangle.

So it seems like the function maps L(\langle1, 0, 0\rangle, \langle0, 1, 0\rangle) = a, L(\langle1, 0, 0\rangle, \langle0, 0, 1\rangle) = b,, and L(\langle0, 1, 0\rangle, \langle0, 0, 1\rangle) = c

I'm not sure how that helps me determine what L(\langle v_1, v_2, v_3\rangle, \langle w_1, w_2, w_3\rangle) is.

Could the function perhaps be L(u, v) = L(\langle v_1, v_2, v_3\rangle, \langle w_1, w_2, w_3\rangle) = a\cdot{}v_1 + b\cdot{}w_3 + c\cdot{}v_2

That seems to satisfy our initial conditions.

My textbook doesn't cover this and my teacher hasn't shown an example of how these functions work, so I'm unsure what to do.

I would start with the definition of "skew-symmetric multilinear function".

The next thing to do would be to note that v = <v1, v2, v3> = v1U1 + v2U2 + v3U3. You can write w in a similar fashion, as a linear combination of U1, U2, and U3.

That's a start...
 
Write v=v1*U1+v2*U2+v3+U3 and similar for w. Use linearity to factor out the vi's and wi's and skew-symmetry to change, for example, <U2,U1> to -<U1,U2>.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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