Independent components of three indexed systems ##T_{ijk}##

[brotherbobby]

Homework Statement

Find the number of independent components of the three-indexed systems below. Assume space to be Euclidean $\text{E}_3$ :

(1) $T_{ijk}$, given that it is $\text{symmetric}$ in its first two indices, i.e. $\underline{T_{ijk}=T_{jik}}$

(2) $T_{ijk}$, given that it is $\text{anti-symmetric}$ in its first two indices, i.e. $\underline{T_{ijk}=-T_{jik}}$

Relevant Equations

(1) In an Euclidean space of dimension $m$ $\,(\text{E}_m)$, an $n$ indexed system will have $n^m$ components in total.

(2) If these indices respect some symmetry due to their mutual change, the number of independent components will reduce.

Attempt :

[The author uses the term "system" with respect to indexed quantities. He reserves the term "tensor" when the components of those quantities respect certain rules when co-ordinates transform.]

Both the "systems" have a total of $3^3 = 27$ components. Of course, they are not all independent.

I calculate the independent components (or otherwise), in a series of possibilities, where one or more of them may be equal.

Let me begin with the first system.

(1) ${\large{T_{ijk}}}\,\text{where}\, \underline{T_{ijk}=T_{jik}}$ :

(a) $T_{iii} = \text{(itself)}\,\text{(all indices equal)}\,$ : Clearly, these values can take any numbers and hence they are independent. Hence, the number of independent components here $\rightarrow 3\quad{\color{green}\checkmark}$.

(b) $T_{iij} = \text{(itself)},\text{(first two indices equal)}\,$ : Clearly, these values can take any numbers and hence they are independent. Hence, the number of independent components here $\rightarrow ^3\!\!P_2 = 6\quad{\color{green}\checkmark}$.

(c) $T_{iji}=T_{jii}\,\text{(the last two, or the first and third indices equal)}\,$ : Counting the number of either of them, we find that the number of independent components here $\rightarrow ^3\!\!P_2 = 6\quad{\color{green}\checkmark}$.

(d) $T_{ijk}=T_{jik}\,\text{(all indices unequal)}\,$ : Counting the number of either of them, we find that the number of independent components here $\rightarrow ^3\!\!C_2 = 3\quad{\color{green}\checkmark}$.

Taking all the independent components above (with green checkmark against them), we find that the total number of independent components for this three-indexed system symmetric in its first two indices are $\boxed{n_i = 18}$.


(2) ${\large{T_{ijk}}}\,\text{where}\, \underline{T_{ijk}=-T_{jik}}$ :

(a) $T_{iii} = 0\;\text{(all indices equal)}$ : Clearly, these values are zero due to anti-symmetry and none of them are independent. Hence, the number of independent components here $\rightarrow 0\quad{\color{green}\checkmark}$.

(b) $T_{iij} = 0\;\text{(first two indices equal)}$ : Clearly, these values are zero due to anti-symmetry and none of them are independent. Hence, the number of independent components here $\rightarrow 0\quad{\color{green}\checkmark}$.

(c) $T_{iji}=-T_{jii}\,\text{(the last two, or the first and third indices equal)}\,$ : Counting the number of either of them, we find that the number of independent components here $\rightarrow ^3\!\!P_2 = 6\quad{\color{green}\checkmark}$.

(d) $T_{ijk}=-T_{jik}\,\text{(all unequal)}\,$ : Counting the number of either of them, we find that the number of independent components here $\rightarrow ^3\!\!C_2 = 3\quad{\color{green}\checkmark}$.

Taking all the independent components above (with green checkmark against them), we find that the total number of independent components for this three-indexed system anti-symmetric in its first two indices are $\boxed{n_i = 9}$.

Request : No answers exist. Are my answers correct, and more importantly, my approach? Is there a shorter way to calculate the number of independent components at once, instead of in steps, as I have done above?

I ask, because I am aware that for a symmetric two-indexed system $T_{ij}=T_{ji}$, the number of independent components are $^3P_2=6$, while for the anti-symmetric case where $T_{ij}=-T_{ji}$, the number becomes $^3C_2=3$. Do similar formula using P's and C's exist for systems of indices 3 and more?

Many thanks.

 

[julian]

The approach is thorough, though perhaps a bit more detailed than necessary.

(1) This symmetry applies to $i$ and $j$. For each fixed $k$, $T_{ijk}$ is a symmetric 3×3 matrix in indices $i,j$. The number of independent components in a symmetric 3×3 matrix is 6. Since $k$ can independently take 3 values, we have 3 such symmetric matrices. So the total number of independent components is 3×6=18.

(2) Similar to (1) but now the for each fixed $k$, $T_{ijk}$ is an anti-symmetric 3×3 matrix in indices $i,j$. The number of independent components in an anti-symmetric 3×3 matrix is 3. So the total number of independent components is 3×3=9.

 

[Orodruin]

There is no need to split into cases as you do. The third index is just an additional copy of the 2-index system for every possible value of the index as compared to the basic 2-index systems you mentioned. It follows directly that the symmetric case is 3x6 = 18 and the anti-symmetric 3x3 = 9.

 

[julian]

I am aware that for a symmetric two-indexed system Tij=Tji, the number of independent components are 3P2=6, while for the anti-symmetric case where Tij=−Tji, the number becomes 3C2=3. Do similar formula using P's and C's exist for systems of indices 3 and more?

Say $i=1,2,\cdots n$, and consider the object $T_{i_i i_2 \cdots i_k}$ that is symmetric under the interchange of any pair of indices.

The number of independent components, $N$, is equal to the number degree-$k$ monomials in $n$ variables $x_1, x_2, \dots, x_n$, i.e., expressions of the form

\begin{align*}
x_1^{j_1} x_2^{j_2} \cdots x_n^{j_n} \quad \text{where} \quad j_1+j_2+\cdots+j_n=k
\end{align*}

For example, $k=3$ and $n=4$, the monomial $x_1^2 x_2$ is the same no matter if it came from $x_1 x_1 x_2$ or $x_1 x_2 x_1$ etc, just as $T_{112} = T_{121}$, and so on. Hence the correspondence.

This is a "stars and bars" combinatorics problem where, for example:

\begin{align*}
**|*| | \quad \text{corresponds to} \quad x_1^2 x_2
\end{align*}

and

\begin{align*}
*| | | ** \quad \text{corresponds to} \quad x_1 x_4^2
\end{align*}

Generally, there will be $k$ stars and $n-1$ bars, so:

\begin{align*}
N = \binom{n+k-1}{k}
\end{align*}


Now, consider the totally antisymmetric object $T_{i_1 i_2 \cdots i_k}$, meaning that swapping any pair of indices changes the sign of the component, where $i=1,2, \cdots, n$. It is zero if any two indices are equal. Thus the number of independent components, $N$, equal to the number of ways to choose distinct indices $i_1 < i_2 < \cdots < i_k$. This is the same as selecting $k$ distinct numbers from $\{1, 2, \dots, n\}$, where the order doesn’t matter. Therefore,

\begin{align*}
N =\frac{n(n-1) \cdots (n-k+1)}{k!} = \binom{n}{k}
\end{align*}

 

[brotherbobby]

Thank you @julian , while I am struggling to understand your method, especially that of the "stars and bars", your results are precious enough to note down and commit to memory.
In answer to my question therefore, which you had highlighted, there does exist formulae for the number of independent components of a system $T_{i_1,i_2,i_3,...,i_n}$ symmetric under an interchange of any pair of indices. Of course it won't work for my problem above though because my system is $T_{ijk}$ symmetric under only the first pair of indices.
If a $k-\,\text{indexed}\,$ system was entirely anti-symmetric, your formula is invaluable. The number of independent components is simply $n\choose k$. So for a three-indexed totally antisymmetric quantity $T_{ijk}$, the number of independent components is $\binom{3}{3}=1$, something I have always done the long way.

 

[julian]

The "stars and bars" thing: Consider the example where $T_{ijk}$ is symmetric under the interchange of any pair of indices, and the indices take the values $1,2,3,4$. For example, all these are the same:

\begin{align*}
T_{112} = T_{121} = T_{211} .
\end{align*}

These can all be represented this by $T_{112}$. Here’s another example:

\begin{align*}
T_{144} = T_{414} = T_{441}.
\end{align*}

These can all be represented this by $T_{114}$. Note we are choosing the component where the values of the indices don’t decrease as you go from left to right to be the representative.

To help us find the number of independent components, we introduce a simple diagram using stars and bars where the number of stars following a bar tells you how many times that particular index occurred. In our example, we would have the list

\begin{align*}
T_{111} &\Leftrightarrow |_1 *** |_2 |_3 |_4 \\
T_{112} &\Leftrightarrow |_1 ** |_2 * |_3 |_4 \\
T_{113} &\Leftrightarrow |_1 ** |_2 |_3 * |_4 \\
T_{114} &\Leftrightarrow |_1 ** |_2 |_3 |_4 * \\
T_{122} &\Leftrightarrow |_1 * |_2 ** |_3 |_4 \\
T_{123} &\Leftrightarrow |_1 * |_2 * |_3 * |_4 \\
T_{124} &\Leftrightarrow |_1 * |_2 * |_3 |_4 * \\
T_{133} &\Leftrightarrow |_1 * |_2 |_3 ** |_4 \\
T_{134} &\Leftrightarrow |_1 * |_2 |_3 * |_4 * \\
T_{144} &\Leftrightarrow |_1 * |_2 |_3 |_4 ** \\
T_{222} &\Leftrightarrow |_1 |_2 *** |_3 |_4 \\
T_{223} &\Leftrightarrow |_1 |_2 ** |_3 * |_4 \\
T_{224} &\Leftrightarrow |_1 |_2 ** |_3 |_4 * \\
T_{233} &\Leftrightarrow |_1 |_2 * |_3 ** |_4 \\
T_{234} &\Leftrightarrow |_1 |_2 * |_3 * |_4 * \\
T_{244} &\Leftrightarrow |_1 |_2 * |_3 |_4 ** \\
T_{333} &\Leftrightarrow |_1 |_2 |_3 *** |_4 \\
T_{334} &\Leftrightarrow |_1 |_2 |_3 ** |_4 * \\
T_{344} &\Leftrightarrow |_1 |_2 |_3 * |_4 ** \\
T_{444} &\Leftrightarrow |_1 |_2 |_3 |_4 ***
\end{align*}

But we can make this even simpler. We don’t really need to label the bars as 1, 2, 3, and 4, because the position of each bar tells us which number it’s for. Also, the first bar isn’t needed: we drop it and the if there are any stars before our new first bar, we know those are 1s. So for example, the first four can be rewritten:

\begin{align*}
T_{111} &\Leftrightarrow *** | | | \\
T_{112} &\Leftrightarrow ** | * | | \\
T_{113} &\Leftrightarrow ** | | * | \\
T_{114} &\Leftrightarrow ** | | | * \\
T_{122} &\Leftrightarrow * | ** | | \\
\text{etc}
\end{align*}

and another four examples are

\begin{align*}
T_{222} &\Leftrightarrow | *** | | \\
T_{223} &\Leftrightarrow | ** | * | \\
T_{224} &\Leftrightarrow | ** | | * \\
T_{233} &\Leftrightarrow | * | ** | \\
T_{234} &\Leftrightarrow | * | * | * \\
\text{etc}
\end{align*}

Each arrangement of stars and bars corresponds to an independent component of $T_{ijk}$. Therefore, the number of arrangements of the stars and bars tells you the number of independent components of $T_{ijk}$ there are.

So, we just need the number of ways to arrange 3 indistinguishable stars and (4-1) indistinguishable bars in a line. This is given by

\begin{align*}
\frac{(3+4-1)!}{3!(4-1)!} = \binom{4-1+3}{3} = 20
\end{align*}