# Help on contraction

1. Dec 11, 2006

### AiRAVATA

Hello guys. Here I am, bothering all of you again...

I am having troubles proving the following:

For $v \in V$ and $\omega \in \Lambda^k(V)$, show that if $v_1,v_2,...,v_n$ is a basis of $V$ with dual basis $\phi_1,\phi_2,...,\phi_n$ then

$$i_{v_j}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})=\left\{ \begin{array}{ll} 0 & j\neq \hbox{ any } i_\alpha \\ (-1)^{\alpha-1} \phi_{i_1}\wedge ... \wedge \hat{\phi}_{i_\alpha}\wedge...\wedge \phi_{i_k} & \hbox{if }j=i_\alpha \right.$$

where $i_v \omega (v_1,...,v_{k-1})=\omega(v,v_1,...,v_{k-1})$ and $\hat{\phi}_{i_\alpha}$ means we are omitting that term.

Use this result to show that for $\omega_1 \in \Lambda^k(V)$ and $\omega_2\in \Lambda^l(V)$ we have
$$i_{v}(\omega_1 \wedge \omega_2)=(i_v\omega_1)\wedge \omega_2+(-1)^k\omega_1\wedge(i_v\omega_2).$$

Here is what I don't understand. In order to prove the first result, I apply the right side to $(v_1,...,v_{j-1},v_{j+1},...,v_k)$, so
$$i_{v_j}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})(v_1,...,v_{j-1},v_{j+1},...,v_k)=(-1)^{j-1}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})(v_1,...,v_j,...,v_k)=\sum_{\sigma \in S} (-1)^\sigma \sigma (\phi_{i_1}\otimes ...\otimes\phi_{i_k}) (v_1,...,v_j,...,v_k)$$

I'm stuck at this point. I know I need to evaluate the tensor product, but I'm going nowhere. Please someone enlighten me...

Last edited: Dec 11, 2006
2. Dec 12, 2006

### AKG

I don't think you mean to define iv on (v1, ..., vk-1) only (since that is just a list of the first k-1 basis vectors of V), but for any k-1 vectors in V, so you probably want to call them w1, ..., wk-1, i.e. you want to define:

$$i_v \omega (w_1, \dots ,w_{k-1}) = \omega (v,w_1,\dots ,w_{k-1})$$

To prove the first result, go back to the definitions.

Case 1, $j \neq i_{\alpha }$ for any $\alpha$. Let's denote vj by w0. We want to prove:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} ) = 0$$

This is true iff for all (k-1)-tuples of vectors in V, (w1, ..., wk-1), the following equation holds:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1}) = 0(w_1, \dots , w_{k-1})$$

The right side is equal to 0, obviously. The left side is:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1})$$

$$= (\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$= k! Alt(\phi _{i_1} \otimes \dots \otimes \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$= \frac{k!}{k!} \sum _{\sigma \in S_{\{ 0, 1, \dots , k-1\} }} sgn(\sigma )(\phi _{i_1} \otimes \dots \otimes \phi _{i_k} )(w_{\sigma (0)}, w_{\sigma (1)}, \dots , w_{\sigma (k-1)})$$

Let A denote S{0, 1, ..., k-1}. Then the above equals:

$$\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

Now for each $\sigma$, there is some n such that $\sigma (n) = 0$. For this n, we get:

$$\phi _{i_{n+1}}(w_{\sigma (n)}) = \phi _{i_{n+1}}(w_0) = \phi _{i_{n+1}}(v_j) = \delta _{i_{n+1}j} = 0$$

with the second last equality holding by definition of the action of an element of a dual basis on an element of the original basis (that $\delta$ being the Kronecker $\delta$), and the last equality holding because $j \neq i_{\alpha }$ for all $\alpha$, hence in particular, $j \neq i_{n+1}$.

So for each $\sigma$, one of the factors in the expression

$$\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

will be 0, hence each term of the sum will be 0, hence the sum is 0, as desired.

Case 2, $j = i_{\alpha }$: your turn.

3. Dec 12, 2006

### AiRAVATA

Oh... I see.

In the case $j=i_{n+1}$, for each $\sigma$ there wil be some n such that

$$= (\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$=\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi_{i_n}}(w_{\sigma(n-1)}) \cdot 1 \cdot \phi_{i_{n+2}}(w_{\sigma(n+1)})\dots\phi _{i_k}(w_{\sigma (k-1)})$$

and then, taking the change

$$i(j)=\left\{\begin{array}{ll} j+1 & \hbox{if } j<n \\ j & \hbox{if } j>n\end{array}\right.$$

the equality becomes

$$=\sum _{\sigma \in A'} sgn (i \circ \sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi_{i_n}}(w_{\sigma(n)}) \phi_{i_{n+2}}(w_{\sigma(n+1)})\dots\phi _{i_k}(w_{\sigma (k-1)})$$

$$=(-1)^{n} \frac{k!}{k!} \sum_{\sigma \in S_{\{1,2,...,k-1\}}} sgn(\sigma)(\phi_{i_1}\otimes\dots\otimes \phi_{i_{n}}\otimes\phi_{i_{n+2}}\otimes \dots \otimes \phi_{i_{k-1}})(w_{\sigma(1)},...,w_{\sigma(k-1)})$$

$$=(-1)^{n}(\phi_{i_1}\wedge \dots \wedge \hat{\phi}_{i_{n+1}}\wedge \dots \wedge \phi_{i_{k-1}})$$

What do you think?

I am guessing that in order to apply this result to the next identity I must use linearity.

Last edited: Dec 12, 2006
4. Dec 12, 2006

### AKG

Your proof ends up being okay, but the way it's written makes it hard to see how you've justified some of the lines. In the second line, it's technically okay to sum over all $\sigma \in A$, but it looks strange, because most of those terms are zero. Also, it kind of makes it look as though for all $\sigma$, $\phi _{i_n+1}(w_{\sigma (n)}) = 1$. Your permutation i doesn't even have a value at n (although the only remaining choice is i(n) = 0, which ends up being the correct choice). Even though $sgn (i\circ \sigma ) = sgn (\sigma \circ i)$, you really should write $sgn (\sigma \circ i)$ because that's the actual permutation you're applying to the subscripts of the w's. Your second last line goes from summing over A to summing over Sk-1. It's not immediately clear why this is justified. Essentially, it is saying that you can sum over this smaller set because for permtuations $\sigma \in A$ such that $\sigma (0) \neq 0$, the term

$$(\phi_{i_1}\otimes\dots\otimes \phi_{i_{n}}\otimes\phi_{i_{n+2}}\otimes \dots \otimes \phi_{i_{k-1}})(w_{\sigma(1)},...,w_{\sigma(k-1)})$$

is zero, and Sk-1 consists precisely of those elements in A which fix 0. But why is the above term 0 if $\sigma$ doesn't fix 0? The way you've written the whole proof, I can't clearly explain why this would be the case.

I would do it as follows:

Case 2, $j = i_{\alpha }$:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1})$$

$$= \sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$ (as computed in post #2)

Define $\tau = (0\, 1\, 2\, \dots \, a-2\, a-1)$. Then the above is:

$$= \sum _{\sigma \in A} sgn (\sigma \circ \tau )\phi_{i_1}(w_{\sigma \circ \tau (0)}) \dots \phi _{i_k}(w_{\sigma \circ \tau (k-1)})$$

$$= \sum _{\sigma \in A} sgn (\sigma )sgn(\tau )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha }}(w_{\sigma (0)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha }}(w_{\sigma (0)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in S_{k-1}} sgn (\sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha -1}}(w_{\sigma (\alpha - 1)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in S_{k-1}} sgn (\sigma )\phi_{i_1} \otimes \dots \otimes \hat{\phi _{i_{\alpha }}} \otimes \dots \otimes \phi _{i_k}(w_{\sigma (1)},\, \dots ,\, w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\phi_{i_1} \wedge \dots \wedge \hat{\phi _{i_{\alpha }}} \wedge \dots \wedge \phi _{i_k}(w_1,\, \dots ,\, w_{k-1})$$

so in this case 2,

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} ) = (-1)^{\alpha - 1}\phi_{i_1} \wedge \dots \wedge \hat{\phi _{i_{\alpha }}} \wedge \dots \wedge \phi _{i_k}$$

as desired.

Last edited: Dec 12, 2006
5. Dec 12, 2006

### AiRAVATA

Thanks a lot for your help.