# What is the use of the Welch lower bound when k > 1?

The Welch lower bound on the maximum cross-correlation (inner product) of $m$ vectors, each of $n$ elements, is given by $c_{max} = f(m, n, k)$ (eg, see Wikipedia Welch bounds). It is always assumed that $m \gt n$ (otherwise the bound is 0). Apparently the Welch bound has been widely applied in the restricted case that the vectors are real-valued and $k = 1.\;$ In that case, $c_{max} = \left[ (m-n)/(n(m-1)) \right]^{1/2},\;$ and is always real. The function has practical use in allowing us to compare the $c_{max}\;$ calculated from any given set of vectors with the theoretical minimum possible value.

In contrast, when $k > 1,\;$ then $c_{max}\;$ is frequently a complex quantity, even if the vectors are real-valued. $\;c_{max}$ becomes real only if $m >> n\;,$ or more specifically, when $m \gt \binom{n+k-1}{k}.\;$
It is unclear to me whether $c_{max}$ has any use when $k \gt 1.$

$\quad$ Q1: Does $k$ have any physical interpretation?

$\quad$ Q2: Is the Welch formula with $k \gt 1\;$ useful in any area of physics, computational science,
$\quad\quad$ electrical engineering, or telecommunications? How do you interpret complex returned values?

Given any collection of $m$ unit-norm (or unit-Euclidean-length) vectors in a real or complex $n$-dimensional vector space $\mathbb R^n$ or $\mathbb C^n$, let $c_{\max}$ be the maximum magnitude of the inner product between two vectors in the set. How small can $c_{\max}$ be? One answer to this question is given by the Welch bounds which are a collection of lower bounds on the minimum possible value of $c_{\max}$. These bounds apply to all collections of $m$ unit-norm vectors in $n$-dimensional space, and they even hold for the case when the collection is a multiset (and thus two vectors are identical, and so $c_{\max}$ necessarily equals $1$).
The Welch bounds are of the form $$c_{\max}^2 \geq \sqrt[k]{C_k(m,n)}, ~k = 1, 2, 3, \ldots \tag{1}$$ or more correctly $$c_{\max}^2 \geq \sqrt[k]{\max\{C_k(m,n), 0\}},~ k = 1, 2, 3, \ldots \tag{2}$$ because the function $C_k(m,n)$ can have negative value. Note for example, that for $k = 1$, $(1)$ becomes $$c_{\max}^2 \geq \frac{m-n}{n(m-1)}$$ where the right side is negative when $m < n$ and $0$ when $m = n$. This is as it should be because we know that in a $n$-dimensional space, we can always find $m \leq n$ orthonormal vectors and so $c_\max^2 = 0$ for this set. Thus, $(2)$ which becomes $$c_{\max}^2 \geq \max\left\{\frac{m-n}{n(m-1)}, 0\right\}\tag{3}$$ is a much better way of expressing the bound. When $m \gg n$ and so $(m-n)/(m-1)$ is approximately $1$, then (as Welch pointed out as also have you in your question) the right side of $(3)$ approaches $n^{-1}$ as $m \to \infty$ and so we have that $$c_\max \geq \frac{1}{\sqrt{n}}\tag{4}$$ as a limiting form of the lower bound.
Turning to the question asked about Welch bounds for $k \geq 2$, these are useful when $m \gg n$ and give larger lower bounds than the right side of $(4)$. If I remember correctly, the Welch bound for $k = 2$ gives something like $$c_\max \geq \frac{2^{1/4}}{\sqrt{n}}$$ when $m \geq \frac{n^3+n^2}{2}$ which is an improvement on $(4)$.