lecture20

# lecture20 - Lecture 20 We begin with a review of last...

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Unformatted text preview: Lecture 20 We begin with a review of last lecture. Consider a vector space V . A tensor T L k is decomposable if T = 1 k , i L 1 = V . A decomposable tensor T is redundant of i = i +1 for some i . We define k k = ( V ) = Span { redundant k-tensors } . (4.94) I I Because I k k , we can take the quotient space L k k k = k ( V ) = L / I , (4.95) defining the map : L k k . (4.96) We denote by A k ( V ) the set of all alternating k-tensors. We repeat the main theorem from last lecture: Theorem 4.27. The map maps A k bijectively onto k . So, A k k . = It is easier to understand the space A k , but many theorems are much simpler when using k . This ends the review of last lecture. 4.6 Wedge Product Now, let T 1 I k 1 and T k 2 . Then T 1 T 2 and T 2 T 1 are in I k , where k = k 1 + k 2 . 2 L The following is an example of the usefulness of k . Let i k i , i = 1 , 2. So, i = ( T i ) for some T i L k i . Define k = k 1 + k 2 , so T 1 T k . Then, we define 2 L ( T 1 T 2 ) = 1 2 k . (4.97) Claim. The product i 2 is well-defined. Proof. Take any tensors T i L k i with ( T i ) = i . We check that ( T 1 T 2 ) = ( T 1 T 2 ) . (4.98) We can write T = T 1 + W 1 , where W 1 I k 1 , (4.99) 1 T = T 2 + W 2 , where W 2 I k 2 . (4.100) 2 Then, T 1 T 2 = T 1 T 2 + W 1 T 2 + T 1 W 2 + W 1 W 2 , (4.101) I k so 1 2 ( T 1 T 2 ) = ( T 1 T 2 ) . (4.102) (4....
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lecture20 - Lecture 20 We begin with a review of last...

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