solutions07

solutions07 - Problem Set #7 Solutions Clifford Algebras...

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Unformatted text preview: Problem Set #7 Solutions Clifford Algebras For the first 4 problems, let R be a ring with 1 and let M n ( R ) be the ring of n n matrices with entries in R . The set R n of column vectors of length n is then an ( M n ( R ) , R )-bimodule, and its transpose t R n is an ( R, M n ( R ))- bimodule. 1. Show that any A M n ( R ) can be written as a sum of rank 1 matrices. Let A M n ( R ) and let E ij M n ( R ) denote the matrix with a 1 in the ( i, j )th slot and zeros elsewhere. Clearly the R-dimension of the image of each E ij acting on R n on the left or on t R n on the right is 1, so in any case E ij is a rank 1 matrix. Letting a ij R be the ( i, j )th entry of A we have A = X 1 i,j n a ij E ij . 2. Given A M n ( R ) , show that there exists v 1 , . . . , v k R n and w t 1 , . . . , w t k t R n such that M = 1 i k v i w t i . Let A M n ( R ) with entries a ij R and let e i R n have a 1 in the i th slot and zeros elsewhere. Then since e i e t j = E ij it follows from part 1 that A = X 1 i,j n a ij e i e t j . Taking v j = i a ij e i and w j = e j we obtain our desired result. 3. Conclude that R n R t R n = M n ( R ) , an isomorphism of ( M n ( R ) , M n ( R ))- bimodules. Note that the e i , 1 i n , form an R-basis for R n , so R n R t R n has basis e i e t j , 1 i, j n . Define : R n R t R n M n ( R ) by set- ting ( e i e t j ) = e i e t j = E ij on this basis and extending ( R, R )-bilinearly to R n R t R n . Let A M n ( R ) with entries a s` R ; then by part b, ( Ae i e t j ) = 1 s,` n a s` e s e t ` e i e t j = 1 s,` n a s` `,i e s e t j = 1 s n a si e s e t j = 1 s n a si ( e s e t j ) = 1 s n a si E sj = 1 s,` n a s` `,i E sj = 1 s,` n a s` E s` E ij = 1 s,` n a s` E s` ( e i e t j ) = A ( e i e t j ) , 1 and by symmetry ( e i e t j A ) = ( e i e t j ) A . We check is well-defined: Let r, a i , b j R . Then using exercise IV.6.5 and the ( R, R )-linearity of we find i ( a i e i ) r j e t j b j = i,j a i ( ( re i ) e t j ) b j = i,j a i ( r ( e i e t j )) b j = i,j a i r ( e i e t j ) b j = i,j a i rE ij b j = i,j a i E ij rb j = i,j a i ( e i e t j ) rb j = i,j a i (( e i e t j ) r ) b j = i,j a i ( e i ( e t j r )) b j = i a i e i r j e t j b j ....
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solutions07 - Problem Set #7 Solutions Clifford Algebras...

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