# I Vector Space of Alternating Multilinear Functions ...

1. Mar 17, 2019 at 2:36 AM

### Math Amateur

I am reading the book: Multivariable Mathematics by Theodore Shifrin ... and am focused on Chapter 8, Section 2, Differential Forms ...

I need some help in order to fully understand the vector space of alternating multilinear functions ...

The relevant text from Shifrin reads as follows:

In the above text from Shifrin we read the following:

" ... ... In particular, if $T \in {\bigwedge}^k ( \mathbb{R}^n )^{ \ast }$, then for any increasing $k$-tuple $I$, set $a_I = T( e_{ i_1} , \cdot \cdot \cdot , e_{ i_k} )$. Then we leave it to the reader to check that

$T = \sum_{ i \text{ increasing } }a_I \text{dx}_I$

... ... ... "

Can someone please help me to prove/demonstrate that $T = \sum_{ i \text{ increasing } }a_I \text{dx}_I$ ... ...

Help will be much appreciated ...

Peter

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In case someone needs access to the text where Shifrin defines the terms of the above post and explains the notation, I am providing access to the start of Chapter 8, Section 2.1 as follows:

Hope that helps ...

Peter

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Last edited: Mar 17, 2019 at 3:07 AM
2. Mar 17, 2019 at 3:49 AM

### andrewkirk

We are told that the set of $d\mathbf x_I$ with $I$ increasing spans the vector space $\Lambda^k(\mathbb R^n)^*$ of alternating multilinear functions from $(\mathbb R^n)^k$ to $\mathbb R$. So for any $T$ in that space there exists a set of real numbers $A_T$ indexed by the set $S$ of increasing $k$-tuples in $\{1, ...,n\}^k$, such that:

$$T=\sum_{J\in S} a_J d\mathbf x_J$$

where $a_J$ is the element of $A_T$ indexed by $k$-tuple $J$.

Applying both sides to the $k$-tuple of vectors $(\mathbf e_{i_1},...,\mathbf e_{i_k})$, for increasing $k$-tuple $I=(i_1,...,i_k)$, we get:

\begin{align*}
T(\mathbf e_{i_1},...,\mathbf e_{i_k})
&=
\sum_{J\in S} a_J d\mathbf x_J (\mathbf e_{i_1},...,\mathbf e_{i_k})
\\&=
\sum_{I\in S} a_J \delta^{i_1,...,i_k}_{j_1,...,j_k}
\end{align*}

by the presumed definition of $d\mathbf x_J$ (not supplied in OP). All the Kronecker deltas in terms in that sum are zero except the one where $J=I$, where the delta is 1. So the sum on the RHS equals $a_I$, giving us the desired result.

3. Mar 17, 2019 at 4:37 AM

### Math Amateur

Hi Andrew ...

Shifrin defines $\text{dx}_I$ as follows ... ...

... Now if $I = (i_1, \cdot \cdot \cdot , 1_k )$ is an ordered $k$-tuple, define

$\text{dx}_I : \mathbb{R}^n \times \cdot \cdot \cdot \mathbb{R}^n \to \mathbb{R}$ by (note $k$ input factors)

$\text{dx}_I ( v_1, \cdot \cdot \cdot , v_k ) = \begin{vmatrix} \text{dx}_{ i_1} ( v_1) & \cdot \cdot \cdot & \text{dx}_{ i_1} ( v_k) \\ \cdot & \cdot \cdot \cdot & \cdot \\ \cdot & \cdot \cdot \cdot & \cdot \\ \text{dx}_{ i_k} ( v_1) & \cdot \cdot \cdot & \text{dx}_{ i_k} ( v_k) \end{vmatrix}$

I think that that would give the result you state ...

Peter

4. Mar 17, 2019 at 4:47 AM

### Math Amateur

Hi Andrew ...

Still reflecting on what you have written ...

Indeed ... you write:

" ... ... So for any $T$ in that space there exists a set of real numbers $A_T$ indexed by the set $S$ of increasing $k$-tuples in $\{1, ...,n\}^k$ ... .. "

Can you explain in simple terms what you mean by this ,,, perhaps also giving a simple example ...

Also ... what is the exact form and nature of the $a_J$ .... ?

Peter

5. Mar 17, 2019 at 5:19 AM

### andrewkirk

Consider where n=5 and k=3 then $A_T$ is the set of all increasing 3-tuples out of the numbers 1,...,5, which is:

123, 124, 125, 134, 135, 145, 234, 235, 245, 345

The set $A_T$ has those ten members. The fourth member is the 3-tuple (1,3,4). If we use J to denote that member then we have $J=(1,3,4)$ and $j_1=1,j_2=3,j_3=4$.

The set $A_T$ doesn't depend much on what T is. All it uses from T is the value of $k$. Given a $k$-tensor T, the set $A_T$ is the set of all increasing k-tuples out of the numbers 1...n. In this example, the tensor T has order 3, and $A_T$ is the set of all 3-tuples (triples) from 1,...,5.

6. Mar 17, 2019 at 5:31 AM

### Math Amateur

Thanks Andrew ... but ... I'm lost ... I'm probably trying you patience ,,,

But can you explain from the definition of the space $\Lambda^k(\mathbb R^n)^*$ how we end up considering a set a set of real numbers $A_T$ in the first place ...

My apologies for being slow to catch on ....

Peter

7. Mar 17, 2019 at 1:58 PM

### Math Amateur

Hi Andrew ... I should be more specific ...

... indeed ... I should ask a specific question ...

... so ...

Now ... $\Lambda^k(\mathbb R^n)^*$ is the vector space of alternating tensors or alternating multilinear mappings of degree or rank $k$ ... and I am assuming that it is a vector space over the field $\mathbb{R}$ ... is that assumption correct?

The basis of $\Lambda^k(\mathbb R^n)^*$ is the set of $k$-linear covectors or mappings ...

$\text{dx}_{I_1}, \text{dx}_{I_2}, \text{dx}_{I_3}, \cdot \cdot \cdot , \text{dx}_{I_l}$ where $l = \begin{pmatrix} n \\ k \end{pmatrix}$

... so an element $T$ of $\Lambda^k(\mathbb R^n)^*$ can be expressed as

$T = a_1 \text{dx}_{I_1} + a_2 \text{dx}_{I_2} + \cdot \cdot \cdot + a_l \text{dx}_{I_l}$ ... where $a_i \in \mathbb{R}$ for all $i$ such that $1 \le i \le l$ ... ...

... ... is that correct?

So I am wondering how the $a_i$ above relate to your $a_J$ ... indeed are the $a_J$ just real numbers ...

Hope you can help ...

Peter

8. Mar 17, 2019 at 4:16 PM

### andrewkirk

That is a spanning set, not a basis, because it includes $d\mathbf x_I$ for non-increasing $I$. So for $k=3$ it would include $d\mathbf x_{1,3,5}$ and $d\mathbf x_{3,1,5}$, which are linearly dependent since $d\mathbf x_{1,3,5}=-d\mathbf x_{3,1,5}$.

To make it a basis, we need to restrict to $d\mathbf x_I$ for increasing $I$.

That is correct if you use the numbers $1,...,m$ to index $A_T$, where $m$ is the number of distinct, increasing $k$-tuples that can be drawn from $1,...,n$. In my example above with k=3,n=5, we have m=10. But you also need to specify an indexing of $A_T$, ie label its elements with the numbers 1 to m.

My approach avoids having to do that, as well as simplifying the notation, by noting that indexes don't have to be integers. We can use the k-tuples themselves to index the k-forms. That is what the author has done above when he writes things like $d\mathbf x_{2,4,5,1}$. That is for the case k=4 and$n\ge 5$, and the example k-tuple is (2,4,5,1). The corresponding coefficient would be $a_{(2,4,5,1)}$.

A longer, but more explanatory presentation of the formula [EDIT: fixed an error in that, which used $I$ instead of $J$ for the index]:

$$T = \sum_{J\in A_T} a_J d\mathbf x_J$$

is

$$T = \sum_{\substack{(i_1,...,i_k)\in \{1,...,n\}^k\\i_1<i_2<....<i_k}} a_{(i_1,...,i_k)} d\mathbf x_{(i_1,...,i_k)}$$

or, even more lengthily:

$$T = \sum_{i_1=1}^n \sum_{i_2=i_1+1}^n...\sum_{i_k=i_{k-1}+1}^n a_{(i_1,...,i_k)} d\mathbf x_{(i_1,...,i_k)}$$

It may help to read the short wiki article on index sets. We are first introduced to indexing in cases where the index set is always a contiguous set of natural numbers starting with 1 or 0. But as we move into more advanced maths, it becomes useful to realise that an indexing of a set is just a surjection from another set (the index set) onto the first one, and the index set doesn't have to be integers. It can for instance be k-tuples. Doing this allows simplification of notation and avoids begging questions such as 'how do we order and number the set $A_T$ of increasing k-tuples?'.

Last edited: Mar 17, 2019 at 5:45 PM
9. Mar 17, 2019 at 5:03 PM

### Math Amateur

Andrew ... thank you for a most helpful post ...

Have gotten the main ideas now ...

Still reflecting on what you have written ...

Thanks again...

Peter