If we take the direct sum of two q<\/em>-matroids $M_1=(E_1,r_1)$ and $M_2=(E_2,r_2)$, it makes sense for $M=M_1\\oplus M_2$ to have as a ground space the direct sum of vector spaces $E_1\\oplus E_2$. It is also reasonable to ask that $M|_{E_1}=M\/E_2=M_1$ and $M|_{E_2}=M\/E_1=M_2$. But then it gets difficult. Where for sets, we can write any subset $A\\subseteq E_1\\sqcup E_2$ as $(A\\cap E_1)\\sqcup(A\\cap E_2)$, such a thing is not true for vector spaces. Suppose for example $E_1=\\langle100\\rangle$ and $E_2=\\langle010,001\\rangle$ (over my favourite finite field $\\mathbb{F}_2$) such that $E_1\\oplus E_2$ is a 3-dimensional space. Then the space $A=\\langle111\\rangle$ does not intersect $E_1$ nor $E_2$. What is its rank going to be?<\/p>\n\n\n\n One could hope that maybe the rank axioms of q<\/em>-matroids take care of this. Look for example at the direct sum of $U_{1,1}$ and $U_{1,2}$ over $\\mathbb{F}_2$.<\/p>\n\n\n\n We see that the interval between $0$ and $\\langle100\\rangle$ and the interval between $\\langle010,001\\rangle$ and $E$ are coloured as $U_{1,1}$, showing $M|_{E_1}=M\/E_2=M_1$. On the other hand, the interval between $0$ and $\\langle010,001\\rangle$ and the interval between $\\langle100\\rangle$ and $E$ are coloured as $U_{1,2}$, showing that $M|_{E_2}=M\/E_1=M_2$. There is now only one way to colour the rest of the lattice:<\/p>\n\n\n\n Unfortunately, this construction becomes not unique already in dimension 4. If we try to make the direct sum $U_{1,2}\\oplus U_{1,2}$, semimodularity of the rank function gives that all 1-dimensional spaces have rank 1 and all 3-dimensional spaces have rank 2. Any 2-dimensional space intersecting $E_1$ or $E_2$ is a basis. But there are 2-dimensional spaces that intersect neither $E_1$ nor $E_2$ (over $\\mathbb{F}_2$, there are three of them) and those can be either a basis or a circuit. Any choice will produce another q<\/em>-matroid.<\/p>\n\n\n\n The solution to this problem is to write the direct sum in a very convoluted way that is not helpful for matroids over sets, but that does have a nice q<\/em>-analogue. This is the following. As mentioned, let $E=E_1\\oplus E_2$. Make a matroid $M_1’$ by adding loops to $M_1$ until its groud space is $E$: $M_1’$ is the q<\/em>-matroid on $E$ such that $M_1’|_{E_1}=M_1$ and $M_1’|_{E_2}$ consists of only loops. Similarly, let $M_2’$ be the q<\/em>-matroid on $E$ such that $M_2’|_{E_2}=M_2$ and $M_2’|_{E_1}$ consists of only loops. Now the direct sum $M_1\\oplus M_2$ is defined as the matroid union $M_1’\\vee M_2’$.<\/p>\n\n\n\n It turns out that adding loops is, inductively, possible in the q<\/em>-analogue. Also matroid union (as an honour to the name of this blog, surely!) has a well-defined q<\/em>-analogue: this operation is defined for matroids on the same ground set, and can thus be generalised to q<\/em>-matroids with the same ground space without the troubles that come with the direct sum of vector spaces. The rank function of the direct sum of two q<\/em>-matroids is \\[ r_{M_1\\oplus M_2}(A)=\\min_{X\\subseteq A}\\{r_{M_1′}(X)+r_{M_2′}(X)+\\dim A-\\dim X\\}. \\]<\/p>\n\n\n\n An equivalent way to phrase this definition of the direct sum of q<\/em>-matroids, is by requiring the direct sum to to be the “most independent” q<\/em>-matroid satisfying $M|_{E_1}=M\/E_2=M_1$ and $M|_{E_2}=M\/E_1=M_2$. This can be made precise in category theory language.<\/p>\n\n\n\n It is unclear what the independent spaces of the direct sum of two q<\/em>-matroids are. The vector space sum of an independent space in $M_1$ and an independent space in $M_2$ is independent in the direct sum. But there are many more independent spaces. Many other properties of the direct sum of matroids fail to have a nice q<\/em>-analogue. What does work in the q<\/em>-analogue, is that the cyclic flats of the direct sum are exactly the sums of the cyclic flats of $M_1$ and $M_2$.<\/p>\n\n\n\n You might have guessed that if $M_1$ and $M_2$ are q<\/em>-matroids represented by matrices $G_1$ and $G_2$ (I’m going to use this notion without a formal definition), then their direct sum is not<\/em> necessarily represented by the matrix \\[ \\left[ \\begin{array}{cc} G_1 & 0 \\\\ 0 & G_2 \\end{array} \\right]. \\] Interestingly, the question about representability can be translated to the language of linear sets<\/em> in finite geometry, making it possible to use several results in finite geometry for q<\/em>-matroids.<\/p>\n\n\n\n [AJNZ26<\/a>] Gianira Alfarano, Relinde Jurrius, Alessandro Neri and Ferdinando Zullo. Representability of the direct sum of uniform q-matroids.<\/em> Combinatorial Theory, 6 (2026).<\/p> [CJ24<\/a>] Michela Ceria and Relinde Jurrius. The direct sum of q-matroids.<\/em> Journal of Algebraic Combinatorics, 59, pp. 291-330 (2024).<\/p> [GLJ23<\/a>] Heide Gluesing-Luerssen and Benjamin Jany. Coproducts in categories of q-matroids.<\/em> European Journal of Combinatorics 112, 103733 (2023).<\/p> [GLJ24<\/a>] Heide Gluesing-Luerssen and Benjamin Jany. Decomposition of q-matroids using cyclic flats.<\/em> SIAM Journal on Discrete Mathematics 38, pp 2940–2970 (2024).<\/p> [GLJ25<\/a>] Heide Gluesing-Luerssen and Benjamin Jany. Representability of the direct sum of q-matroids.<\/em> Journal of Algebraic Combinatorics 61, 51 (2025).<\/p><\/p>\n","protected":false},"excerpt":{"rendered":" One of the most straightforward operations you can do when you want to make a new matroid out of old ones, is taking the direct sum. The direct sum of the matroids $M_1=(E_1,r_1)$ and $M_2=(E_2,r_2)$ is the matroid $M$ on … Continue reading ![\"\"](\"https:\/\/matroidunion.org\/wp-content\/uploads\/2026\/08\/U11U12half-1024x514.png\")
<\/a><\/figure>\n\n\n\n![\"\"](\"https:\/\/matroidunion.org\/wp-content\/uploads\/2026\/08\/U11U12full-1024x514.png\")
<\/a><\/figure>\n\n\n\nReferences<\/h3>\n\n\n\n