I'm working on some questions in tropical geometry, and my problem led me to create the following generalization of a determinant:


Let $A$ be an $m \times n$ matrix with $m \le n$, and positive integer multiplicities $m\_i$ assigned to the rows with $\sum\_{i=1}^m m\_i =n$ so $A$ is "square with multiplicity". I consider labeled partitions $\mathcal I= \{I\_1, \dots, I\_m\}$ of $\{1, \dots, n\}$ with $|I\_i| = m\_i$. Then the "determinant with multiplicity" is
$$
\sum\_{\mathcal I} \operatorname{sgn}(\mathcal I)\prod\_{i=1}^m \prod\_{k\in I\_i} A\_{ik}.
$$
Notice that when $m=n$ (so all $m\_i=1$) then this is just the usual determinant. 


I actually haven't thought about what $\operatorname{sgn}(\mathcal I)$ should be in general, because I created that definition by lifting from a tropical definition-- I call a matrix "tropically non-singular with multiplicity" if
$$
 \max\_{\mathcal I} \left\{\sum\_{i} \sum\_{k \in I\_i} A\_{ik}\right\},
$$
is achieved exactly once.


I expect "tropically non-singular with multiplicity" to have some nice properties generalizing those of the usual (Strumfels) tropical non-singularity. Some of those proofs use a lift to Puiseaux series and then the properties of the usual determinant, so if I want to imitate those proofs, I would need to understand the properties of the "determinant with multiplicity".


So my question is: does anybody recognize the definitions above as something that has been studied before? If so, can you point me to a reference?


**EDIT**


Some more motivation: As a function, the second display above is the same as the tropical determinant/permanent of the actual square matrix formed by repeating the $i$th row $m\_i$ times. I prefer my version because it eliminates repeated terms that "come only from" the repeated rows, that is, a matrix with a repeated row is not automatically considered to be singular. I guess this is funny classically, but I think I have a tropical geometric interpretation that makes sense.



Has this generalization of a determinant (assigning multiplicities to the rows) been studied?
**Summary**


There is a clear connection to the [permanent](https://en.wikipedia.org/wiki/Permanent), which, if I understand correctly, yields the same tropicalization, and a more speculative connection to the determinant. The reason is that OP's formula can be rewritten as an [immanant](https://en.wikipedia.org/wiki/Immanant_of_a_matrix)-like expression evaluated on an auxiliary matrix $\widehat{A}$ obtained by repeating the rows of $A$ according to the multiplicities $m\_i.$ For arguably the most natural choice of the coefficients in the expansion, where they are all equal $1$, one obtains the *permanent* of $\widehat{A}$ (up to a constant multiple). On the other hand, one can also choose the coefficients for an arbitrary $m\leq n$ in a way that reproduces the determinant when $m=n.$ I don't know whether this leads to a meaningful generalization of the determinant.




---


**Analysis**


Modulo the choice of the coefficients, the procedure outlined in the question amounts to the following.



> 
> 1. Replace the rectangular $m\times n$ matrix $A$ with the square $n\times n$ matrix $\widehat{A}$ by repeating the $i$th row of $A$ $m\_i$ times.
> 2. Consider the row expansion of the "immanant"
> $\sum\_{\sigma\in S\_n}k\_{\sigma}\prod\_{i=1}^n \widehat{A}\_{i,\sigma(i)},$ where $k\_\sigma$ are some coefficients that need to be specified ("the signs").
> 
> 
> 


The partition $\mathcal{I}=\{I\_1,I\_2,\ldots,I\_m\}$ of the set $\{1,2,\ldots,n\}$ of the column indices from the original formula is $\large\{\sigma$(first $m\_1$ indices), $\sigma$(next $m\_2$ indices), $\ldots$, $\sigma$(last $m\_m$ indices)$\large\}$ and the coefficient $\operatorname{sgn}(\mathcal{I})=\sum k\_\sigma,$ with the sum taken over all permutations $\sigma$ corresponding to that partition $\mathcal{I}.$


In the case that $k\_\sigma=1$ for all $\sigma\in S\_n,$ we get the permanent of $\widehat{A},$ which is equal to the constant $(\prod\_{i=1}^{m}m\_i!)$ times (OP's expression for all coefficients equal to $1$). Although this is non-standard in the context of permanents, the constant can be eliminated by setting $k\_{\sigma}=0$ for permutations $\sigma$ that "scramble" some of the parts $I\_k$, cf the next paragraph.


One can also consider the following "determinant" choice of $k\_\sigma.$ It is defined to be $\operatorname{sgn}(\sigma)$ if the descent set of $\sigma$ is contained in $\{m\_1,m\_1+m\_2,\ldots,m\_1+\ldots+m\_{m-1}\}$ and $0$ otherwise. In other words, for each partition $\mathcal{I}$ consider the element $\sigma\_{\mathcal{I}}\in S\_n$ which maps the first $m\_1$ indices to $I\_1$ *preserving the order*, the next $m\_2$ indices to $I\_2$ *preserving the order*, and so on, and set $\operatorname{sgn}(\mathcal{I})=\operatorname{sgn}(\sigma\_{\mathcal{I}}).$ The resulting expression is always non-trivial and for $m=n,$ it yields $\det(A).$ On the other hand, if $m<n$ then the matrix $\widehat{A}$ contains repeating rows and hence its determinant is $0.$




---


**Examples.**


1. For $m=1$, the matrix $A$ is a single row, $A=[a\_1,a\_2,\ldots,a\_n],$ the multiplicity $m\_1=n,$ there is only one possible partition, where $I\_1=\{1,2,\ldots,n\},$ and (assuming the coefficient is $1$) the OP's expression is $\prod\_{i=1}^n a\_i,$ that is the product of the entries of $A.$ The matrix $\widehat{A}$ is the row $[a\_1,a\_2,\ldots,a\_n]$ repeated $n$ times and its permanent is $n!$ times the product of the entries of $A.$ Even though the "determinant" choice of the "sign" coefficient is $1,$ it appears more natural to view the product of the entries of $A$ as an analogue of the permanent rather than the determinant. On the other hand, unless $n=1,$ the determinant of $\widehat{A}$ is $0$ (corresponding to the "sign" coefficient being $0$).
2. For $m=2,$ the matrix $A$ has two rows and $n$ columns and the multiplicity is a pair $(p,q)$ of natural numbers that add up to $n.$ A partition $\mathcal{I}=\{I,J\}$ of the set of column indices with $|I|=p$ and $|J|=q$ leads to the choice of $p$ entries in the first row of $A$ and $q$ entries in the second row in such a way that every column contains exactly one chosen entry. OP's procedure is to multiply the chosen entries together and take a linear combination of these products over different choices with to-be-specified coefficients $\operatorname{sgn}(I,J).$ It is easy to see that this is effectively the same as repeating the first row of $A$ $p$ times and the second row $q$ times to form a square matrix $\widehat{A}$ and computing its "immanant" as above. There are two natural choices for the coefficients: $\operatorname{sgn}(I,J)=1$, leading to the permanent of $\widehat{A},$ and $\operatorname{sgn}(I,J)=\operatorname{sgn}(\sigma),$ where $\sigma$ is the *shortest* permutation corresponding to the partition $(I,J)$, namely $\sigma(i)=I\_i$ for $1\leq i\leq p$ and $\sigma(p+i)=J\_{i}$ for $1\leq i\leq q$ (it is assumed that the parts $I$ and $J$ are represented by *increasing* sequences of indices).
3. For $m=n,$ the multiplicities are necessarily all equal to $1$ and partition $\mathcal{I}$ is the same as a permutation $\sigma$ of $\{1,2,\ldots,n\}.$ In this case, the recipe amounts to the usual row expansion of the "immanant" of $A$, which gives the determinant or the permanent if the coefficients $k\_\sigma$ are $\operatorname{sgn}\sigma$ or $1,$ respectively. Of course, there are also other possibilities, for example, $k\_{\sigma}=\chi\_{\lambda}(\sigma),$ where $\chi\_{\lambda}$ is an irreducible character of the symmetric group $S\_n,$ corresponds to the immanant $\operatorname{Imm}\_{\lambda}(A)$ of $A.$


