Kronecker delta versus identity matrix

Question

How should $\delta_k^j$ be regarded? Is it a scalar that takes on variable values? A 3x3 identity matrix (in 3 dimensions)?

The wikipedia article on raising and lowering indices with the metric tensor includes the phrase:

"where $\delta_{ik}$ is the Kronecker delta or identity matrix''

however I think this is wrong.

Consider this: $$ \delta_k^j a_j = \delta_k^1 a_1 + \delta_k^2 a_2 + \delta_k^3 a_3 = a_k $$ In the pattern above $\delta$ is not equivalent to the identity matrix. Rather, it is a selection vector, i.e., $$ \delta_k^j a_j = [0, 1, 0] \left[ \begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\right] $$ where the 1 is appearing in the $k$th position (here k=2). However, as a $(\,_1^1)$ tensor it seems like it should be equivalent to a matrix

Answer 1

When you write $\delta_k^j a_j = \delta_k^1 a_1 + \delta_k^2 a_2 + \delta_k^3 a_3 = a_k$ it is essentially acting as the identity matrix, it's just renaming the index from j to k. It just seems weird because of the summation convention.

Answer 2

The $n\times n$ identity matrix is sometimes written as $(\delta_i^j)_{1\leq i,j\leq n}$, so you identify the matrix with its entries which are Kronecker deltas... in this case it's a bit confusing.

Answer 3

For a given pair of values $i$ and $j$, the symbol $\delta_{ij}$ is just a number. If $i=j$, then the number is $1$, otherwise, it is $0$.

The identity matrix can be simply written as $I=(\delta_{ij})_{i,j=1}^n$, that is why wikipedia calls the identity matrix "a Kroenecker delta matrix". It may be strangely written, but not horrible.

Answer 4

Sometimes it helps to unpackage the index notation shorthand and explicitly write tensors and vectors in terms of a basis.

Let $e_1, e_2, \ldots$ be a set of basis vectors, and let $e^1, e^2, \ldots$ be the associated dual basis.

The Kronecker delta can be understood in terms of linear operator $\delta$ such that

$${\delta_i}^j = e^i[\delta(e_j)]$$

We have the prescription that ${\delta_i}^j = 1$ only when $i=j$. Otherwise, it is zero.

Let's observe now the action of $\delta$ on a vector $a = a^1 e_1 + a^2 e_2 + \ldots = \sum_j a^j e_j$. Set this equal to some vector $b$ like so:

$$b = \delta(a) = \sum_j a^j \delta(e_j)$$

Now, we don't have any idea what $\delta(e_j)$ should be, but we know what happens if we act on $b$ with an arbitrary dual basis vector $e^k$:

$$e^k(b) = b^k = \sum_j a^j e^k[\delta(e_j)]$$

But we know that $e^k[\delta(e_j)]$ is nonzero only when $k = j$. Therefore, we can remove the summation and set all references to $j$ to $k$ instead:

$$e^k(b) = b^k = a^k e^k[\delta(e_k)] = a^k \times 1 = a^k$$

It bears saying that $k$ is an arbitrary index. It is simply a label we give to some whole number $1, 2, \ldots, n$ in our n-dimensional space. That we choose to give this label the symbol $k$ should not be interpreted as having meaning. We could just as easily call this label "banana". What is important is that we've established, for any choice of $k$, $b^k = a^k$.

So now, more concretely, we can now recover an expression for $b$ in terms of the basis vectors:

$$\delta(a) = b = \sum_k b^k e_k = \sum_k a^k e_k = a$$

Therefore, since $\delta(a) = a$ for any $a$, we conclude that $\delta$ is the identity map.

Answer 5

Ok, I now see the answer to my question. Thank you for the answers.

In my words: Going back to the two equations I wrote in the question, they are true for any particular $k$, so the delta does not act as an identity matrix when evaluated for a particular $k$. But I have to remember that $k$ is a free index, so it should take on all values, and then the delta is like an identity.