à la Riesz


sabiendo que b^i=g^{si}b_s y que \beta^i(b_j)={\delta^i}_j tanto como \langle b^i,b_j\rangle={\delta^i}_j, entonces

\beta(\quad)=\langle b^i,\quad\rangle

o bien

\beta(X)=\langle b^i,X\rangle

para todo X\in V

 

Ahora para un covector arbitrario f\in V^* se tiene:

f(\quad)

f(X)=f(X^sb_s)=X^sf(b_s)   ————–  (A)

v.s.

f(b_s)\beta^s(\quad)

f(b_s)\beta^s(X)=f(b_s)\langle b^s,X\rangle

=f(b_s)\langle b^s,X^tb_t\rangle

=f(b_s)X^t\langle b^s,b_y\rangle

=f(b_s)X^t{\delta^s}_t

=f(b_s)X^s   —————  (B)

/..\!\!\cdot   (A,B)

f(\quad)=f(b_s)\beta^s(\quad)
=f(b_s)\langle b^s,\quad\rangle
=\langle f(b_s)b^s,\quad\rangle

o bien

f(X)=f(b_s)\beta^s(X)
=f(b_s)\langle b^s,X\rangle
=\langle f(b_s)b^s,X\rangle

para todo X\in V.

Es decir el representante del covector fà la Riesz- es:

f(b_s)b^s

esto es una combinación lineal en la base recíproca b^i de V, which represent the basic covectors \beta^i

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Filed under algebra, cucei math, differential geometry, mathematics, multilinear algebra, what is math, what is mathematics

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