solutions07

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

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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 φ : R n ⊗ R t R n → M n ( R ) by set- ting φ ( 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, φ ( Ae i ⊗ e t j ) = φ ∑ 1 ≤ s,` ≤ n a s` e s e t ` e i ⊗ e t j = φ ∑ 1 ≤ s,` ≤ n a s` δ `,i e s ⊗ e t j = φ ∑ 1 ≤ s ≤ n a si e s ⊗ e t j = ∑ 1 ≤ s ≤ n a si φ ( e s ⊗ e t j ) = ∑ 1 ≤ s ≤ n a si E sj = ∑ 1 ≤ s,` ≤ n a s` δ `,i E sj = ∑ 1 ≤ s,` ≤ n a s` E s` E ij = ∑ 1 ≤ s,` ≤ n a s` E s` φ ( e i ⊗ e t j ) = Aφ ( e i ⊗ e t j ) , 1 and by symmetry φ ( e i ⊗ e t j A ) = φ ( e i ⊗ e t j ) A . We check φ is well-defined: Let r, a i , b j ∈ R . Then using exercise IV.6.5 and the ( R, R )-linearity of φ we find φ ∑ i ( a i e i ) r ⊗ ∑ j e t j b j = ∑ i,j a i φ ( ( re i ) ⊗ e t j ) b j = ∑ i,j a i φ ( r ( e i ⊗ e t j )) b j = ∑ i,j a i rφ ( 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 φ ( e i ⊗ e t j ) rb j = ∑ i,j a i φ (( e i ⊗ e t j ) r ) b j = ∑ i,j a i φ ( e i ⊗ ( e t j r )) b j = φ ∑ i a i e i ⊗ r ∑ j e t j b j ....
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solutions07 - Problem Set#7 Solutions Clifford Algebras For...

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