Vector Space of Alternating Multilinear Functions ....

In summary: I_m}## ...No problem Peter.Yes, you are correct, ##\Lambda^k(\mathbb{R}^n)^*## is a vector space over the field ##\mathbb{R}##.Yes, the basis is the set of all ##k##-linear covectors or mappings ##d\mathbf{x}_I##.Now, for any ##k##-tensor ##T## in this space, there exists a set of real numbers ##A_T## indexed by the set of increasing ##k##-tuples in ##\{1, ..., n\}^k##. This is because any ##k##-tensor can be written as a linear combination
  • #1
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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:
?temp_hash=c5a985b43e67ff351bbb3452d56f5618.png


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
==========================================================================================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:
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?temp_hash=a6ac9c21b636d70cd5c23e172af350a3.png

?temp_hash=a6ac9c21b636d70cd5c23e172af350a3.png

Hope that helps ...

Peter
 

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  • #2
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.
 
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  • #3
Hi Andrew ...

Thanks for your help ...

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
andrewkirk said:
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.
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
Math Amateur said:
" ... ... 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## ... ?
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.
 
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  • #6
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
Math Amateur said:
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
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
Math Amateur said:
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}##
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##.

... 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?
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 realize 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?'.
 
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  • #9
Andrew ... thank you for a most helpful post ...

Have gotten the main ideas now ...

Still reflecting on what you have written ...

Thanks again...

Peter
 

1. What is a vector space of alternating multilinear functions?

A vector space of alternating multilinear functions is a mathematical structure that consists of a set of functions that take multiple vector inputs and produce a scalar output. These functions are linear, meaning that they follow the properties of linearity, and are alternating, meaning that they change sign when the order of the inputs is switched. This vector space is often denoted as Λ(V), where V is a vector space.

2. What are the applications of vector spaces of alternating multilinear functions?

Vector spaces of alternating multilinear functions have various applications in mathematics, physics, and engineering. They are commonly used in differential geometry, where they are used to define the exterior derivative and the wedge product. They are also used in physics to describe quantities such as torque and angular momentum. In engineering, they are used in the study of stress and strain in materials.

3. How are vector spaces of alternating multilinear functions different from other vector spaces?

Vector spaces of alternating multilinear functions are different from other vector spaces in that they are defined by the properties of linearity and alternativity, rather than by a specific set of vectors. This means that the elements of this vector space are not vectors themselves, but rather functions that operate on vectors. Additionally, the operations of addition and scalar multiplication are defined differently in this vector space compared to others.

4. What is the dimension of a vector space of alternating multilinear functions?

The dimension of a vector space of alternating multilinear functions is equal to the number of vector inputs that the functions take. For example, if the functions take two vector inputs, the dimension of the vector space would be two. This is because the number of vector inputs determines the number of basis elements needed to span the vector space.

5. How are vector spaces of alternating multilinear functions related to tensors?

Vector spaces of alternating multilinear functions are closely related to tensors, as they are both used to represent multilinear functions. Tensors are more general than vector spaces of alternating multilinear functions, as they can also represent non-linear functions. However, when the inputs and outputs of a tensor are all vectors, it can be represented as a vector space of alternating multilinear functions. In this way, vector spaces of alternating multilinear functions can be seen as a special case of tensors.

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