16
$\begingroup$

My question basically is, is Kronecker delta $\delta_{ij}$ or $\delta^{i}_{j}$. Many tensor calculus books (including the one which I use) state it to be the latter, whereas I have also read many instances where they use the former. They cannot be the same as the don't have the same transformation laws. What I think is that since

$\delta_{j}^{i}$ = ($\delta_{j}^{i}$)$'$, but $\delta_{ij}$ doesn't, so the latter cannot be a tensor. But the problem is that both have the same value :- ($1,0$) depending upon the indices. So it makes me think that $\delta_{ij}$ is just the identity matrix $I$ and not a tensor, and $\delta^{i}_{j}$ is a function. But since $\delta_{j}^{i}$ also has the same output as $\delta_{ij}$, WHAT IS THE DIFFERENCE?

I think it could be matrix representations. IN GENERAL, is there a difference between matrix representations of $\delta_{ij}$, $\delta^{ij}$ and $\delta_{j}^{i}$ (or any other tensor for that matter). Please answer these (the difference between mixed indices AND matrix representations).

$\endgroup$
16
$\begingroup$

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.]

II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing

$$\tag{1} V \otimes V^{*}~\stackrel{\delta}{\longrightarrow}~ \mathbb{F}$$

$$ v\otimes f ~\stackrel{\delta}{\mapsto}~ f(v) , \quad v\in V, \quad f\in V^{*},$$

between an $\mathbb{F}$-vector space $V$ and its dual vector space $V^{*}$.

III) If we choose a basis $(e_i)_{i\in I}$ for $V$, there is an dual basis $(e^{j*})_{j\in I}$ for $V^{*}$ such that

$$\tag{2} e^{j*}(e_i)~=~\delta^j_i~:=~\left\{ \begin{array}{rcl} 1 &\text{for}& i=j, \\ \\ 0 &\text{for}& i\neq j. \end{array}\right. $$

[Here we distinguish between covariant and contravariant indices.] Then for a vector $v=\sum_{i\in I} v^i e_i\in V$ and a covector $ f=\sum_{j\in I} f_j e^{j*}\in V^{*}$, the contraction map (1) is

$$\tag{3} \delta(v,f)~=~ \sum_{i,j\in I} f_j \delta^j_i v^i~=~ \sum_{i\in I}f_i v^i. $$

In other words, $\delta^j_i$ is the matrix representation for the $\delta$ contraction map (1). It's an interesting fact that the matrix representation is independent of the choice of basis $(e_i)_{i\in I}$ for $V$, as long as we choose the corresponding dual basis for $V^{*}$ in the natural way. We often say that $\delta^j_i$ transforms as a tensor, or is a tensor.

IV) Now what about $\delta_{ij}$ with lower indices? Well, first we must introduce a symmetric bilinear form, or metric,

$$\tag{4} V\times V ~\stackrel{g}{\longrightarrow}~ \mathbb{F} $$ $$ g(v,w)=g(w,v) .$$

If we choose a basis $(e_i)_{i\in I}$ for $V$, then we can write

$$ \tag{5} g ~=~ \sum_{i,j\in I} g_{ij}~ e^{i*}\otimes e^{j*} .$$

Often we will choose a metric which is the unit matrix in a certain basis

$$\tag{6} g_{ij} ~=~\delta_{ij}~:=~\left\{ \begin{array}{rcl} 1 &\text{for}& i=j, \\ \\ 0 &\text{for}& i\neq j. \end{array}\right. $$

If we now choose another basis then the matrix representation $g_{ij}$ for the metric (4) will in general change. It will in general no longer be the unit matrix $\delta_{ij}$. We say that $\delta_{ij}$ does not transform as a tensor under general change of bases/coordinates.

In a nutshell, $\delta_{ij}$ with lower indices implicitly signals the presence of a metric (4), or in other words, a notion of length scale in the vector space $V$. It is important to realize that the choice of a metric (4) in $V$ is a non-canonical choice.

V) However, once we are given a metric $g$, it is natural to study changes of bases/coordinates that preserve this metric $g$. These correspond to orthogonal transformations and $\delta_{ij}$ behaves as a covariant tensor under such orthogonal transformations.

$\endgroup$
  • $\begingroup$ Thank you @Qmechanic..your explanation is excellent..upvote!! $\endgroup$ – GRrocks Jun 14 '14 at 10:27
5
$\begingroup$

tl;dr

All of $\delta^{ij}$, $\delta_{ij}$, $\delta^i\,_j$, $\delta_i\,^j$ are different, unless you are working with cartesian tensors (tensors on a Euclidean space represented in cartesian coordinates). With cartesian tensors they are all have the same values, even though they technically mean different things. In the general case $\delta^i\,_j$, $\delta_i\,^j$ still always give you the identity matrix, but $\delta_{ij} = \mathbf{e}_i \cdot \mathbf{e}_j$ which only yields the identity matrix for an orthonormal basis and $\delta^{ij}$ is the matrix inverse of $\delta_{ij}$ ($\delta$ is really the metric tensor).

rant

It is a bad idea to teach/learn tensors calculus by starting with cartesian tensors, which is what I suspect your book is doing, precisely because the key features of tensor calculus become trivial in that setting. Tensor calculus is designed for working with curved spaces and/or curvilinear coordinates. This is where the extra machinery of tensor calculus does real work. Also tensor calculus is primarily designed to deal with fields: in tensor calculus a "vector" is usually a vector field and similarly for "tensor".

To understand the motivations behind tensor calculus it is important to remember that the usual definition of a vector space does not include an inner product. Inner products are additional structure and different inner products can be defined on top of the same underlying vector space. This helps to see the difference between a vector space and its dual, which can otherwise seem trivial. Even when an inner product is defined, the usual component product formula for it is only valid in bases that are othonormal wrt to the product.

It is only in the general setting that an explicit metric tensor is really needed and there is a real difference between contravariant and covariant vector components.

I highly recommend ch 14 of Penrose's Road to Reality for a conceptual explanation of tensors.

$\endgroup$
  • $\begingroup$ Thanx @Daniel Mahler, nice,concise answer...and I have already read that chapter long back.. :) $\endgroup$ – GRrocks Jun 14 '14 at 10:30
2
$\begingroup$

These Kronecker symbols have the same matrix representations, as you said, just the unit matrix. The indices are placed at upper or lower positions to suit Einstein summation convention (http://en.wikipedia.org/wiki/Einstein_notation). They are always used together with covariant and contravariant vectors (http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors) in curvilinear coordinate systems. I learned these things in the context of electromagnetic theory. If you are familiar with electromagnetic theory too, I would recommend the sections 1.14 - 1.17 in the book Electromagnetic Theory by Stratton. There you can find a quite clear explanation.

$\endgroup$
  • $\begingroup$ Thanx @Pu Zhang.. $\endgroup$ – GRrocks Jun 13 '14 at 13:51
1
$\begingroup$

Usually the symbol $ \delta_{ij} $ is used for the delta Kronecker function, while $ \delta_{i}^{j} $ is a $ (1,1) $ tensor. Is well described on Wikipedia: http://en.wikipedia.org/wiki/Kronecker_delta

Sometimes the use is misleading. For example, for the Pauli matrices: $ [\sigma_a, \sigma_b]_{+} = 2 \delta_{a b} $, but more precisely you have to write $ [\sigma_a, \sigma_b]_{+} = 2 \delta_{a b}\,\mathbb{I}_{2} $. Anyway the answer of Pu Zhang is correct.

$\endgroup$
  • 1
    $\begingroup$ But @Rexcircus, both are used in tensor calculus. So what is the ddifference between their applications? Can you give examples of the difference in their usage?? $\endgroup$ – GRrocks Jun 13 '14 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.