# Help on Contraction: Proving i_v(\omega_1 \wedge \omega_2)

• AiRAVATA
In summary: The last equality holds because of the definition of the antisymmetrization operator and the fact that the \phi 's are linearly independent. The third last equality holds because sgn (\sigma ) = sgn (\sigma )sgn (\tau ) when \sigma fixes the subscripts of the w's, and using \
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...

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AiRAVATA said:
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.
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.

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.

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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.

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Thanks a lot for your help.

## 1. What is a contraction?

A contraction is a mathematical operation used to combine two tensor fields into a single tensor field. It involves multiplying one tensor field by the dual of the other, and then summing over a common index.

## 2. How is a contraction written mathematically?

A contraction is written using the symbol $$\cdot$$ or $$\otimes$$ between the two tensor fields, followed by the indices that are being contracted over. For example, $$A \cdot B = A^{\mu \nu} B_{\mu \nu}$$.

## 3. What is the purpose of proving i_v(\omega_1 \wedge \omega_2)?

The purpose of proving i_v(\omega_1 \wedge \omega_2) is to show that the contraction of a vector field v with the wedge product of two covector fields $$\omega_1$$ and $$\omega_2$$ is equal to the wedge product of the contractions of v with each covector field individually. This is an important property in tensor calculus and has many applications in physics and engineering.

## 4. Can you provide an example of proving i_v(\omega_1 \wedge \omega_2)?

Yes, consider a vector field v = (v^1, v^2, v^3) and two covector fields $$\omega_1 = dx^1 + dx^2$$ and $$\omega_2 = dx^2 + dx^3$$. Then, the contraction of v with $$\omega_1 \wedge \omega_2$$ is:

i_v(\omega_1 \wedge \omega_2) = v^1 (\omega_1 \wedge \omega_2)_{12} + v^2 (\omega_1 \wedge \omega_2)_{23} + v^3 (\omega_1 \wedge \omega_2)_{31}

= v^1 (dx^1 \wedge dx^2)_{12} + v^2 (dx^2 \wedge dx^3)_{23} + v^3 (dx^3 \wedge dx^1)_{31}

= v^1 dx^1 + v^2 dx^3 + v^3 dx^2

= $$(v^1 dx^1 + v^2 dx^3) \wedge (v^2 dx^2 + v^3 dx^1)$$

= (i_v \omega_1) \wedge (i_v \omega_2)

## 5. Why is this property important in tensor calculus?

This property is important because it allows for the simplification and manipulation of tensor equations. It also helps in understanding the relationship between different tensor fields and how they interact with each other. Additionally, this property is used in various fields of physics and engineering, such as in the study of fluid dynamics, electromagnetism, and general relativity.

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