# Contraction of a tensor

1. Oct 7, 2007

### joe2317

Let Y$$_{1}$$,..,Y$$_{k}$$ be vector fields and let A be a tensor field of type $$^{k}_{1}$$. Could you explain how applying k contractions to A$$\otimes$$Y$$_{1}$$$$\otimes$$...Y$$_{k}$$ yields A(Y$$_{1}$$...Y$$_{k}$$)?

Actually, could you first explain why contraction of w$$\otimes$$Y is equal to w(Y)?
Here, w is a 1-form and Y is a vector field.
Thank you.

2. Oct 7, 2007

### HallsofIvy

Staff Emeritus
Isn't that pretty much the definition of "contraction"?

3. Oct 7, 2007

4. Oct 7, 2007

### AlphaNumeric2

$$w \otimes Y$$ has incides $$w_{i} \otimes Y^{j}$$. Contract the indices to make $$w \otimes Y$$ into a scalar gives $$w_{i} \otimes Y^{i}$$. This is the definition of w(Y).

Similarly for everything else.

5. Oct 7, 2007

### quetzalcoatl9

a special case is the dot product of two vectors, this is how everyone really things about contraction anyway

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