On the fractional total domatic numbers of incidence graphs

1.   Introduction

Let $H = (V, \mathscr{E})$ be a $hypergraph$, where $V$ is a finite vertex set and $\mathscr{E}$ is a finite edge set such that every edge $e\in \mathscr{E}$ is a subset of $V$. A vertex subset $T\subseteq V$ is called a $transversal$ of $H$, if each edge of $H$ contains at least one vertex of $T$. The $transversal$ $number$ of $H$ is the minimum size among all transversals in $H$, and it is denoted by $\tau(H)$. The $disjoint\ transversal\ number$ of a hypergraph $H$ is the maximum number of disjoint transversals in $H$, and it is denoted by $disj_{\tau}(H)$. Goddard and Henning ^[1] studied the fractional disjoint transversal number. Given a hypergraph $H$ and a family of (not necessarily distinct) transversals $\mathscr{F}$ of $H$, let $r_{\mathscr{F}}$ be the maximum times that any vertex appears in $\mathscr{F}$. The $fractional\ disjoint\ transversal\ number$ is defined as

Goddard and Henning ^[1] gave some bounds of $DT_{f}(H)$.

Lemma 1.1. ^[1] For every isolate-free hypergraph $H$ of order $n$,

The minimum size of all edges of $H$ is called the $anti$-$rank$ of $H$ and denoted by $r(H)$. Goddard and Henning ^[1] showed a lower bound on $DT_{f}(H)$ using the anti-rank of a hypergraph $H$.

Lemma 1.2. ^[1] If a hypergraph $H$ has order $n$ and anti-rank $k$, then

For $e\in \mathscr{E}(H)$, if edge $e$ has size $k$, $e$ is called a $k$-edge. When every edge of $H$ is a $k$-edge, we call $H$ $k$-$uniform$. A $complete$ $k$-uniform hypergraph on $n$ vertices, denoted by $K^{(k)}_{n}$, has all $k$-subsets of $\{1, \cdots, n\}$ as edges. If $D$ is a minimum transversal of $K_n^{(k)}$, then $\tau(K_n^{(k)}) = |D|\geq n-k+1$ because each $k$-subset of $V(K_n^{(k)})$ contains at least one vertex in $D$. By Lemmas 1.1 and 1.2, $DT_{f}(K_n^{(k)}) = n/(n-k+1)$.

Goddard and Henning ^[1] discussed the fractional disjoint transversal number of the disjoint union of hypergraphs.

Lemma 1.3. ^[1] If $H$ is the disjoint union of isolate-free hypergraphs $H_{1}, H _{2}, \cdots, H_{k}$, then

The $degree$ of a vertex $v$ in $H$, denoted by $d(v)$, is the number of edges containing $v$ in $H$. Let $\delta(H)$ and $\Delta(H)$ be the minimum degree and the maximum degree of hypergraph $H$, respectively. If every vertex $v\in V(H)$ has degree $k$, we say that $H$ is a $k$-$regular$ hypergraph. For $k\geq 2$, let $\mathscr{H}_{k}$ denote the class of all $k$-regular $k$-uniform hypergraphs. Henning and Yeo ^[2] showed that if $H\in\mathscr{H}_{3}$, then $DT_{f}(H)\geq2$. They also proved that for all $k\geq 3$, if $H\in\mathscr{H}_{k}$, then $DT_{f}(H)\geq \frac{k}{1+\ln k}$, and this bound is essentially the best possible.

A $polychromatic$ (or $panchromatic$) $m$-$coloring$ of hypergraph $H$ is a mapping $f:V(H)\rightarrow \{1, 2, \cdots, m\}$ such that all $m$ colors appear on each edge in $H$ (see ^[3,4]). In particular, when $m = 2$, it is also called a $2$-coloring of $H$. Obviously, in a polychromatic $m$-coloring of $H$, each color class is a transversal of $H$. So, a hypergraph $H$ has a polychromatic $m$-coloring if and only if $H$ has $m$ disjoint transversals. Let $H$ be a hypergraph with maximum degree $\Delta$ and anti-rank $r$. The $hall\ ratio$ of hypergraph $H$ is defined as $h(H) = \min\{|\cup\mathscr{J}|/|\mathscr{J}|:\emptyset\neq\mathscr{J}\subseteq\mathscr{E}\}$, where $\cup\mathscr{J} = \cup_{J\in\mathscr{J}}J$. Kostochka and Woodall ^[4] showed that, if (ⅰ) $h(H) > r-1$ or (ⅱ) $r\geq3$, $h(H)\geq r-1$, then $disj_{\tau}(H)\geq r$. Bollobás et al. ^[3] proved that, for a family $\mathscr{H}$ of hypergraphs with maximum degree $\Delta$ and anti-rank $r$, each $H\in \mathscr{H}$ has $disj_{\tau}(H)\geq r/(\ln\Delta+O(\ln\ln\Delta))$; (ii) for all $\Delta\geq 2$, $r\geq 1$, $\min_{H \in\mathscr{H}}\{disj_{\tau}(H)\}\leq \max\{1, O(r/\ln\Delta)\}$; (ⅲ) for each sequence $\Delta$, $r\rightarrow \infty$ with $r = \omega(\ln\Delta)$, $\min_{H\in\mathscr{H}}{disj_{\tau}(H)}\leq(1+o(1))r/\ln \Delta$. Li and Zhang ^[5] gave a lower bound $disj_{\tau}(H)\geq \lfloor {r}/{\ln (c \Delta r^{2})}\rfloor$, where $0 < c = c(\Delta, r) < 1.5582$. Henning and Yeo ^[6] confirmed that every hypergraph $H$ $\in$ $\mathscr{H}_{k}$ contains $m$ disjoint transversals for $k\geq 2$ and $2\leq m\leq \frac{k}{3\ln k}$. Jiang, Yue and Zhang ^[7] showed that, for a fixed constant $q$, every hypergraph $H$ with $h$ edges and anti-rank $r$ has a polychromatic $m$-coloring such that each color appears at least $q$ times on every edge, where $m\geq {r(1-o(1))}/{(2\ln h)}$.

Let $G$ be a graph. A $total\ dominating\ set$ of $G$ is a vertex set that intersects the neighborhood of every vertex of graph $G$. The $total\ domatic\ number$ of graph $G$ is the maximum number of disjoint total dominating sets of $G$ and denoted by $td(G)$. Construct the $open\ neighborhood\ hypergraph$ $ON(G)$ of $G$ as follows. The vertex set of $ON(G)$ is $V(G)$, and the edges of $ON(G)$ are the open neighborhoods of vertices in $G$. Clearly, a total dominating set of $G$ is a transversal of $ON(G)$. Therefore, $td(G) = disj_{\tau}(ON(G))$. That is to say, graph $G$ has $m$ disjoint total dominating sets if and only if $ON(G)$ has $m$ disjoint transversals. So, the disjoint total dominating set partition problem of graphs corresponds to the disjoint transversal partition problem of a special class of hypergraphs.

An $m$-vertex-coloring of graph $G$ is called a $coupon\ coloring$ (or $thoroughly\ dispersed\ coloring$) if every color appears in every open neighborhood of $G$ (see ^[8]). Obviously, in a coupon coloring of $G$, a color class is a total dominating set, and thus the maximum number $m$ of colors in coupon colorings of $G$ is equivalent to $td(G)$. Chen et al. ^[8] proved every $k$-regular graph $G$ has $td(G)\geq (1-o(1))k/\ln k$. Henning and Yeo ^[6] showed every $k$-regular graph $G$ has $td(G)\geq 2$, provided $k\geq4$. Goddard and Henning ^[1] confirmed that every planar graph has $td(G)\leq 4$, and the bound is tight.

Let $\mathscr{M}$ be a family of (not necessarily distinct) total dominating sets of $G$. Let $r_{\mathscr{M}}$ be the maximum times that any vertex of $G$ appears in $\mathscr{M}$. Let

It is called the $fractional \; total \;domatic\; number$ of $G$ ^[1]. Let $\gamma_{t}$ be the minimum size among all total dominating sets in $G$. Goddard and Henning ^[1] gave lower and upper bounds on the fractional total domatic number of a graph.

Lemma 1.4. ^[1] If $G$ is an isolate-free graph of order $n$, then

Goddard and Henning ^[1] established the relation between the fractional total domatic number of graph $G$ and the fractional disjoint transversal number of its open neighborhood hypergraph $ON(G)$.

Lemma 1.5. ^[1] For every isolate-free graph $G$, $FTD(G) = DT_{f}(ON(G))$.

Goddard and Henning ^[1] showed that every claw-free graph $G$ with $\delta(G)\geq 2$ has $FTD(G)\geq 3/2$. Also, they showed that every planar graph has $td(G)\leq 4$ and $FTD(G)\leq 5-\frac{12}{n}$. Henning and Yeo ^[2] verified a conjecture in ^[1] that every connected cubic graph had $FTD(G)\geq 2$. For a $k$-regular graph $G$, its open neighborhood hypergraph $ON(G)\in \mathscr{H}_k$. By Lemma 1.5 and the conclusion that $DT_{f}(H)\geq \frac{k}{1+\ln k}$ for each $H \in \mathscr{H}_k$ due to Henning and Yeo ^[2], it follows that if $G$ is a $k$-regular graph, then $FTD(G)\geq \frac{k}{1+\ln k}$, where $k\geq3$. For more relevant results, see ^[9].

The $subdivision\ graph$ of graph $G$, denoted by $S(G)$, is the graph obtained from $G$ by subdividing every edge of $G$ exactly once. So, $V(S(G)) = V(G)\bigcup E(G)$. Goddard and Henning ^[1] obtained the following result.

Theorem 1.1. ^[1] For integer $n\geq3$, $FTD(S(K_{n})) = \frac{n}{n-1}$.

The $incidence\ graph$ of hypergraph $H$ is a bipartite graph $I(H) = (X, Y, E)$, where $X = V(H)$, $Y = \mathscr{E}(H)$, $xy\in E$ if and only if $x\in X$, $y\in Y$ and $x\in y$. Note that $S(G)$ is exactly the incidence graph of $G$. Goddard and Henning ^[1] extended Theorem 1.1 to incidence graphs of complete $k$-uniform hypergraphs for any integer $k\geq 3$.

Theorem 1.2. ^[1] Let $k\geq 3, n\geq 2k$ be two integers. Then,

In this paper, we extend Theorem 1.2 by removal of the restriction that "$n\geq 2k$". Moreover, we generalize Theorem 1.2 to the incident graphs of $h$-balanced $n$-partite complete $k$-uniform hypergraphs and obtain a similar result. In Section 2, we discuss and determine the fractional disjoint transversal numbers for $n$-partite complete $k$-uniform hypergraphs. In Section 3, we determine the fractional total domatic numbers for the incidence graphs of balanced $n$-partite complete $k$-uniform hypergraphs.

2.   The fractional disjoint transversal numbers of complete uniform hypergraphs

A hypergraph $H$ is called an $n$-$partite$ $complete$ $k$-uniform hypergraph if $H$ satisfies the following:

(1) $H$ has a vertex set partition $\{V_{1}, V_{2}, \cdots, V_{n}\}$;

(2) $|V_{i}| = h_i \geq 1$, $1\leq i\leq n$;

(3) edge set of $H$ consists of all $k$-tuples, each of which exactly intersects $k$ $V_{i}$s.

We denote an $n$-partite complete $k$-uniform hypergraph by $K^{(k)}_{h_{1}, h_{2}, \cdots, h_{n}}$. The following result shows us that the fractional disjoint transversal number of $K^{(k)}_{h_{1}, h_{2}, \cdots, h_{n}}$ is irrelevant to the indices $h_i, 1\leq i\leq n$.

Theorem 2.1. For positive integers $h_i, n, k$, where $n\geq k$ and $1\leq i\leq n$, $DT_{f}(K^{(k)}_{h_{1}, h_{2}, \cdots, h_{n}}) = \frac{n}{n-k+1}$.

Proof. Define $H = K^{(k)}_{h_{1}, h_{2}, \cdots, h_{n}}$. Let the vertex set partition of $H$ be $V(H) = \{V_{1}, V_{2}, \cdots, V_{n}\}$. For any transversal $F$ of $H$, $V(H)\setminus F$ intersects at most $k-1$ $V_{i}$s. That is to say, $F$ contains at least $n-k+1$ distinct $V_{i}$s. Furthermore, for an arbitrary transversal family $\mathscr{F}$ of $H$, by the Pigeonhole Principle, we have that $r_{\mathscr{F}}\geq\frac{|\mathscr{F}|(n-k+1)}{n}$, implying that $\frac{|\mathscr{F}|}{r_{\mathscr{F}}}\leq\frac{n}{n-k+1}$. So, $DT_{f}(H)\leq\frac{n}{n-k+1}$. Consider a transversal family $\mathscr{F}$, which consists of all distinct unions of any $n-k+1$ different $V_{i}s$. Then, $|\mathscr{F}|$ = $\tbinom{n}{n-k+1}$. Since every vertex of $H$ is in $\tbinom{n-1}{n-k}$ transversals of $\mathscr{F}$, we have $\frac{|\mathscr{F}|}{r_{\mathscr{F}}}$ = $\tbinom{n}{n-k+1}$/$\tbinom{n-1}{n-k} = \frac{n}{n-k+1}$. Consequently, $DT_{f}(H) = \frac{n}{n-k+1}$.

When $h_i = h$ for each $1\leq i\leq n$, $K^{(k)}_{h_{1}, h_{2}, \cdots, h_{n}}$ is called $h$-$balanced$, simply denoted by $K_{h\times n}^{(k)}$.

Corollary 2.1. For positive integers $n, k, h$ with $n\geq k$, $DT_{f}(K_{h\times n}^{(k)}) = \frac{n}{n-k+1}$.

3.   The fractional total domatic numbers of the incidence graphs of balanced complete uniform hypergraphs

Let $H = (V, \mathscr{E})$ be a hypergraph with $V = \{x_1, x_2, \cdots, x_n\}$ and $\mathscr{E} = (E_1, E_2, \cdots, E_s)$. The maximum size of all edges of $H$ is called the $rank$ of $H$ and denoted by $R(H)$. An edge subset $S\subseteq \mathscr{E}$ is called a $cover$ of $H$ if $S$ contains all vertices in $V(H)$. If the edge set of $H$ can be partitioned into $m$ disjoint covers, then $H$ has a cover $m$-decomposition. The maximum number $m$ is denoted by $cd(G)$. The $dual$ of $H$ is denoted by $H^*$, whose vertices $e_1, e_2, \cdots, e_s$ correspond to the edges of $H$ and whose edges are $X_i = \{ e_j| x_i \in E_j, 1\leq j\leq s\}$, $i = 1, 2, \cdots, n$. Thus, $\delta(H) = r(H^{*})$ and $\Delta(H) = R(H^{*})$. Clearly, an $m$-transversal-partition of $H$ corresponds to a cover $m$-decomposition of $H^*$, and $disj_{\tau}(H) = cd(H^*)$.

Goddard and Henning ^[1] referred to the following relationship between $H$ and the open neighborhood hypergraph of $I(H)$. For completeness, we give a proof.

Lemma 3.1. Let $H$ be a hypergraph. Then, $ON(I(H))$ consists of two components, $H$ and its dual $H^{*}$.

Proof. Let $V(H)$ = $\{u_{1}, u_{2}, \cdots, u_{n}\}$. By the definition of incidence graph, the vertex set of $I(H)$ is $V(H)\cup \mathscr{E}(H)$. Then, the $ON(I(H))$ consists of two components: $H_{1}$, which has vertex set $V(H)$, and $H_{2}$, which has vertex set $\mathscr{E}(H)$.

$H_{1}$ is formed by the open neighborhoods of the elements in $\mathscr{E}(H)$. Note that in $I(H)$, for each $e = \{u_{1}, u_{2}, \cdots, u_{s}\}\in \mathscr{E}(H)$, the open neighborhood of $e$ is itself, i.e., $\{u_{1}, u_{2}, \cdots, u_{s}\}$. So, $H_1\cong H$.

In $I(H)$, the open neighborhood of an element $u_{i}\in V(H)$ is the set of incident edges of $u_{i}$, $i = 1, 2, \cdots, n$, which forms an edge of $H_{2}$. For each $u_{i}\in V(H)$, we denote its corresponding edge in $H_{2}$ by $U_{i}$. For an element $e\in V(H_{2}) = \mathscr{E}(H)$ and an element $U_{i}\in E(H_{2})$, $e$ is incident with $U_{i}$ in $H_{2}$ if and only if $e$ is incident with $u_{i}$ in $H$. This means that $H_{2}\cong H^*\cong H_{1}^*$.

Next, we improve Theorem 1.2 by removal of the restriction that "$n\geq 2k$". Let $P = (w_{1}, w_{2}, \cdots, w_{p})$ be a permutation. We call $(w_{j}, w_{j+1}, \cdots,$ $w_{p}, w_{1}, \cdots, w_{j-1})$ the $jth$ $rotation$ of $P$, $1\leq j\leq p$. Clearly, the permutation $P$ has $p$ different rotations.

Theorem 3.1. Let $k, n$ be two integers and $n > k\geq 2$. Then,

Proof. We define $H = K^{(k)}_{n}$. By Lemmas 1.3, 1.5 and 3.1, $FTD(I(H)) = DT_{f}(ON(I(H))) = \min\{DT_{f}(H)$, $DT_{f}(H^{*})\}\leq DT_{f}(H)$. Also, by Corollary 2.1, there is $FTD(I(H))\leq \frac{n}{n-k+1}$.

Next, we prove that the bound is also a lower bound of $FTD(I(H))$ by constructing a special family of total dominating sets.

We know that $I(H)$ is a bipartite graph with $V(I(H)) = V(H)\cup E(H)$. Clearly, $|E(H)| = \binom{n}{k}$. Let $V(H) = \{v_1, v_2, \cdots, v_n\}$. Assume that $n = ak+b$, where $a\geq 1, \ 0\leq b < k$. Given a permutation $(v_1, v_2, \cdots, v_n)$, we pick a subset $F_1\subset V(I(H))$ following three rules, as below:

$({\rm{R1}})$ $A_1 = \{v_1, v_2, \cdots, v_{n-k}, v_n\}$;

$({\rm{R2}})$ $B_1 = \{e_0^1, e_1^1, e_2^1, \cdots, e_a^1\}$,

where $e_0^1 = \{v_{n-k+1}, \cdots, v_n\}$ and

$e_j^1 = \{v_{kj-k+1}, \cdots, v_{kj}\}$, $1\leq j\leq a$;

$(R3)$ $F_1 = A_1\cup B_1$.

Note that $e_0^1 = e_a^1$ if and only if $b = 0$. Since each element of $E(H)$ has $k$ neighbors, and there are exactly $k-1$ vertices of $V(H)$ not in $F_1$, each element of $E(H)$ is dominated by $F_1$ in $I(H)$. Also, by $\cup_{j = 0}^a e_j^1 = V(H)$, each element of $V(H)$ is dominated by $F_1$. That is to say, $F_1$ is a total dominating set of $I(H)$.

Let $(u_1^i, u_2^i, \cdots, u_n^i) = (v_i, v_{i+1}, \cdots, v_n, v_1, \cdots, v_{i-1})$ denote the $i$th rotation of the sequence $(v_1, v_2, \cdots, v_n)$. Analogously, based on the sequence $(u_1^i, u_2^i, \cdots, u_n^i)$, we can obtain a total dominating set $F_i = A_i\cup B_i$ of $I(H)$ for each $1\leq i\leq n$. It is easy to see that each element of $V(H)$ appears exactly $n-k+1$ times in the family of total dominating sets $\mathscr{F} = \{F_1, F_2, \cdots, F_n\}$. When $b = 0$, for each $1\leq i\leq n$, $e_1^i, e_2^i, \cdots, e_a^i$ are distinct, and each of them appears exactly $a$ times in $\mathscr{F}$, i.e., $B_h = B_{k+h} = \cdots = B_{(a-1)k+h}$ for each $1\leq h\leq k$. When $b > 0$, for each $1\leq i\leq n$, $e_0^i, e_1^i, e_2^i, \cdots, e_a^i$ are distinct (See Figure 1). Note that, for any $1\leq j\leq n$, the sequential $k$-tuple $\{v_{j}, v_{j+1}, \cdots, v_{j+k-1}\} = e_0^{k+j}$ (both the subscripts and $k+j$ are mod $n$.) This means that $\cup_{i = 1}^n\{e_0^i, e_1^i, e_2^i, \cdots, e_a^i\} = \cup_{i = 1}^n \{e_0^i\}$. It is easy to see that $e_0^1, e_0^2, \cdots, e_0^n$ are distinct. Next, we show $e_0^i$ appears exactly $a+1$ times for each $1\leq i\leq n$.

● When $1\leq i\leq k$, $e_0^i = e_a^{i+b} = e_{a-1}^{i+b+k} = \cdots = e_{1}^{i+b+(a-1)k}$;

● when $tk+1\leq i\leq tk+k$ and $1\leq t\leq a-1$, $e_t^{i-tk} = \cdots = e_1^{i-k} = e_0^i = e_a^{i+b} = e_{a-1}^{i+b+k} = \cdots = e_{t+1}^{i+b+(a-t-1)k}$;

● when $ak+1\leq i\leq n$, $e_0^i = e_a^{i-ak} = \cdots = e_{2}^{i-2k} = e_{1}^{i-k}$.

We give a claim as follows.

Claim. $n-k+1\geq a+1$.

Recall that $n > k\geq 2$. If $a = 1$, then $b\geq 1$. So, $n-k+1 = k+b-k+1\geq 2 = a+1$. If $a\geq 2$, then $n-k+1\geq ak-k+1 = a+a(k-1)-k+1 = a+(a-1)(k-1)\geq a+1$.

In short, $|\mathscr{F}| = n$, and $r_{\mathscr{F}} = n-k+1$. By the definition of the fractional total domatic number, $FTD(I(H))\geq \frac{n}{n-k+1}$.

Remark. According to the definition of $K^{(k)}_{n}$, the condition that $n\geq k$ is fundamental. The case that $n = k$ will be discussed in Theorem 3.2, and the case that $n > k = 1$ is contained in Theorem 3.3. In either of the cases, $FTD(I(K^{(k)}_{n})) = 1.$

The following Theorems 3.2, 3.3 determine the fractional total domatic number of $I(K_{h\times n}^{(k)})$ and generalize Theorem 1.2. Before that, we give a relation between a hypergraph and its sub-hypergraph on the fractional disjoint transversal number. For two hypergraphs $H$ and $\widehat{H}$, we call $\widehat{H}$ a $sub$-$hypergraph$ of $H$ if $V(\widehat{H})\subseteq V(H)$ and $\mathscr{E}(\widehat{H})\subseteq \mathscr{E}(H)$.

Lemma 3.2. Let $\widehat{H}$ be a sub-hypergraph of $H$. Then, $DT_{f}(\widehat{H})\geq DT_{f}(H)$.

Proof. Let $DT_{f}(H) = r$. By the definition of the fractional disjoint transversal number, there exists a transversal family $\mathscr{F}$ of $H$ such that $\frac{|\mathscr{F}|}{r_{\mathscr{F}}} = r$. By $V(\widehat{H})\subseteq V(H)$ and $\mathscr{E}(\widehat{H})\subseteq \mathscr{E}(H)$, we can obtain a corresponding transversal family $\mathscr{\widehat{F}}$ of $\widehat{H}$ by removal of the vertices in $V(H)\setminus V(\widehat{H})$ from $\mathscr{F}$. Obviously, $|\widehat{\mathscr{F}}\ | = |\mathscr{F}|$, and $r_{\widehat{\mathscr{F}}}\leq r_{\mathscr{F}}$. So, $\frac{|\widehat{\mathscr{F}}|}{r_{\widehat{\mathscr{F}}}}\geq \frac{|\mathscr{F}|}{r_{\mathscr{F}}} = r$. Then, we have $DT_{f}(\widehat{H})\geq \frac{|\widehat{\mathscr{F}} |}{r_{\widehat{\mathscr{F}}}}\geq r$.

A $matching$ in a hypergraph is a set of non-intersecting edges. The matching number $\mu(H)$ is the size of a largest matching of hypergraph $H$.

Theorem 3.2. Let $n$, $h$ be two positive integers. Then,

Proof. When $n = 1$, $|V(I(K_{h\times 1}^{(1)}))| = 2h$. Clearly, $td(I(K_{h\times 1}^{(1)})) = 1$, and $\gamma_{t}(I(K_{h\times 1}^{(1)})) = 2h$. By Lemma 1.4, we have $FTD(I(K_{h\times 1}^{(1)})) = 1$. Next, we assume that $n\geq 2$.

When $h = 1$, $K_{1\times n}^{(n)}\cong K_{n}^{(n)}$ has only one edge, which contains $n$ vertices. Clearly, $I(K_{n}^{(n)})$ is a star, and $|V(I(K_{n}^{(n)}))| = n+1$. The edge $e$ in $K_{n}^{(n)}$ corresponds to a vertex $e$ in $I(K_{n}^{(n)})$. Every total dominating set must contain vertex $e$ in $I(K_{n}^{(n)})$. Thus, for every family $\mathscr{M}$ of total dominating sets of $I(K_{n}^{(n)})$, vertex $e$ appears $|\mathscr{M}|$ times in $\mathscr{M}$. So, $r_{\mathscr{M}} = |\mathscr{M}|$. By the definition of the fractional total domatic number, $FTD(I(K_{n}^{(n)})) = 1$.

Next, we focus on the cases that $n\geq2$ and $h\geq2$. Define $H = K_{h\times n}^{(n)}$. By Lemma 3.1, $ON(I(H))$ consists of two components $H$ and $H^{*}$. Then, by Corollary 2.1, $DT_{f}(H) = n$. In the following, we will show that $DT_{f}(H^{*})\geq n$. By Lemma 1.3, $DT_{f}(ON(I(H))) = \min\{DT_{f}(H)$, $DT_{f}(H^{*})\}$ = $n$. Also, by Lemma 1.5, $FTD(I(H)) = DT_{f}(ON(I(H))) = n$.

Now, we discuss $DT_{f}(H^{*})$. First, $|V(H)| = nh$, and $|\mathscr{E}(H)| = h^{n}$. Since each edge contributes $n$ to the degree sum of $H$, $\sum_{v\in V(H)} d(v) = h^{n}n$. By $H$ regularity, we have $d_H(v) = \frac{h^{n}n}{nh} = h^{n-1}$ for each $v\in V(H)$. Then, we know that $H^{*}$ is $h^{n-1}$-uniform and has $h^n$ vertices.

Here, we establish a claim.

$Claim$. $\mathscr{E}(H)$ can be partitioned into $h^{n-1}$ disjoint maximum matchings.

Array all vertices of $H$ into a matrix with $h$ rows and $n$ columns as follows, in such a way that the $j$th column consists of the $j$th partite vertices of $V(H)$, and denote the matrix by $A_{1, 1, \cdots, 1}$.

Every row of $A_{1, 1, \cdots, 1}$ corresponds to an edge in $H$. The set of all rows of $A_{1, 1, \cdots, 1}$ corresponds to a matching of $H$. Clearly, it is a maximum matching of $H$, and $\mu(H) = h$. We use $A_{1, i_2, i_3, \cdots, i_n}$ to denote the matrix obtained from $A_{1, 1, \cdots, 1}$ by replacing the $j$th column $(v_{1j}, v_{2j}, \cdots, v_{hj})^T$ with its $i_j$th rotation, where $2\leq j \leq n$, $1\leq i_j\leq h$. Then, we can obtain $h^{n-1}$ distinct matrices, which correspond to $h^{n-1}$ maximum matchings. We show that arbitrary two of such matchings are disjoint. Pick arbitrarily two distinct matrices $A = A_{1, i_2, \cdots, i_n}$ and $B = A_{1, i_2', \cdots, i_n'}$. Without loss of generality, assume $i_j\neq i_j'$. For the $s$th row of $A$ and the $t$th row of $B$, when $s\neq t$, the first elements are different; when $s = t$, the $j$th elements are different. That is to say, the corresponding matchings of $A$, $B$ are disjoint. Recalling that $|\mathscr{E}(H)| = h^n$, we partition $\mathscr{E}(H)$ into $h^{n-1}$ disjoint maximum matchings. The claim is proved.

Note that every maximum matching in $H$ corresponds to a part of $V(H^{*})$. Thus, $H^{*}$ has $h^{n-1}$ parts, each of which contains $h$ vertices. It is easy to see that $H^{*}$ is $h^{n-1}$-partite $h^{n-1}$-uniform and non-complete. We can extend $H^{*}$ into a $K_{h \times h^{n-1}}^{(h^{n-1})}$ by adding the missing $h^{n-1}$-edges. By Corollary 2.1, we have $DT_{f}(K_{h \times h^{n-1}}^{(h^{n-1}\;\;\;)}\;\;) = h^{n-1}$. By Lemma 3.2, $DT_{f}(H^{*})\geq DT_{f}({K_{h \times h^{n-1}}^{(h^{n-1}\;\;\;)}}\;\;\;) = h^{n-1}$. Noting that $h^{n-1}\geq n$ when $h\geq2$, we have $DT_{f}(H^{*})\geq n$ and then complete the proof.

Theorem 3.3. Let $n$, $k$, $h$ be three positive integers and $n > k$. Then, $FTD(I(K_{h\times n}^{(k)})) = \frac{n}{n-k+1}$.

Proof. Define $H = K_{h\times n}^{(k)}$. When $k = 1$, $|V(I(H))| = 2nh$. Clearly, $td(I(H)) = 1$ and $\gamma_{t}(I(H)) = 2nh$. By Lemma 1.4, we have $FTD(I(H)) = 1$. When $h = 1$, it is done by Theorem 3.1. In the following, we assume that $h\geq 2$, $k\geq 2$.

By Lemma 3.1, $ON(I(H))$ consists of two components, $H$ and $H^{*}$. By Corollary 2.1, $DT_{f}(H) = \frac{n}{n-k+1}$. If $DT_{f}(H^*)\geq \frac{n}{n-k+1}$, then $FTD(I(K_{h\times n}^{(k)})) = DT_f(ON(I(H))) = \min\{DT_{f}(H), DT_{f}(H^{*})\} = \frac{n}{n-k+1}.$ By $FTD(I(K_{n}^{(k)})) = \frac{n}{n-k+1}$, we know that $DT_{f}((K_{n}^{(k)})^*)\geq \frac{n}{n-k+1}$. Next, we show that $DT_{f}(H^*)\geq \frac{n}{n-k+1}$ for $h\geq 2$ and $k\geq 2$. Before that, we need some properties of $K_{n}^{(k)}$ and its dual.

Let $V(K_{n}^{(k)}) = \{v_{1}, v_{2}, \cdots, v_{n}\}$ and $E(K_{n}^{(k)}) = \{E: E\subset V(K_{n}^{(k)}), |E| = k\} = \{E_i: 1\leq i\leq \binom{n}{k}\}$. By the duality, we know that $V((K_{n}^{(k)})^*) = \{e_i: 1\leq i\leq \binom{n}{k}\}$, $E((K_{n}^{(k)})^*) = \{V_{1}, V_{2}, \cdots, V_{n}\}$, and $e_i$ is incident with $V_{j}$ if and only if $E_i$ is incident with $v_{j}$, i.e., $e_i\in V_{j}$ if and only if $v_{j}\in E_i$ for all $1\leq i\leq \binom{n}{k}, \ 1\leq j\leq n$. Pick a family of (not necessarily distinct) transversals $\mathscr{F}$ of $(K_{n}^{(k)})^*$ such that $DT_{f}((K_{n}^{(k)})^*) = \frac{|\mathscr{F}|}{r_{\mathscr{F}}}$. According to the definition of the dual of a hypergraph, we can establish the following observation.

Observation. Let $F = \{e_{1}, e_{2}, \cdots, e_{s}\}\subseteq V((K_{n}^{(k)})^*)$ and $f = \{E_{1}, E_{2}, \cdots, E_{s}\}\subseteq E(K_{n}^{(k)})$. Then, the following statements are equivalent.

(1) $F$ is a transversal of $(K_{n}^{(k)})^*$;

(2) $F\cap V_j\neq \emptyset$ for each element $V_j\in E((K_{n}^{(k)})^*)$;

(3) for each element $V_j\in E((K_{n}^{(k)})^*)$, there exists an element $e_{i_j}\in F$ such that $e_{i_j}\in V_j$;

(4) for each element $v_j\in V(K_{n}^{(k)})$, there exists an element $E_{i_j}\in f$ such that $v_j\in E_{i_j}$;

(5) $\cup_{i = 1}^s E_i = V(K_{n}^{(k)}) = \{v_{1}, v_{2}, \cdots, v_{n}\}$;

(6) $f$ is a cover of $K_{n}^{(k)}$.

Assume that the $n$ parts of $K_{h\times n}^{(k)}$ are $X_1, X_2, \cdots, X_n$, where $X_p = \{v_p^1, v_p^2, \cdots, v_p^h\}$ for each $1\leq p\leq n$. We can partition $V(K_{h\times n}^{(k)})$ into $h$ subsets $Y_1, Y_2, \cdots, Y_h$ such that $Y_q = \{v_{1}^q, v_{2}^q, \cdots, v_{n}^q\}$ for each $1\leq q\leq h$. We next give a family of transversals of $(K_{h\times n}^{(k)})^*$ based on $\mathscr{F}$. For each $F\in \mathscr{F}$, we construct a corresponding transversal $F(h)$ of $(K_{h\times n}^{(k)})^*$ as follows. Without loss of generality, assume that $F = \{e_{1}, e_{2}, \cdots, e_{s}\}$. Then, $f = \{E_{1}, E_{2}, \cdots, E_{s}\}$ is a cover of $K_{n}^{(k)}$, i.e., $\cup_{i = 1}^s E_i = \{v_{1}, v_{2}, \cdots, v_{n}\}$. Noting that $E_i \in E(K_{n}^{(k)})$, we may assume that $E_i = \{v_{i_1}, \cdots, v_{i_k}\}$ for each $1\leq i\leq s$. Set $f^q = \{E_{1}^q, E_{2}^q, \cdots, E_{s}^q\}$, where $E_i^q = \{v_{i_1}^q, \cdots, v_{i_k}^q\}$, $1\leq i\leq s, 1\leq q\leq h.$ Then, $\cup_{i = 1}^s E_i^q = \{v_{1}^q, v_{2}^q, \cdots, v_{n}^q\} = Y_q$. Define

It follows that $f(h)$ is a cover of $K_{h\times n}^{(k)}$ because $\cup_{E\in f(h)} E = \cup _{q = 1}^h \cup_{i = 1}^s E_i^q = \cup _{q = 1}^h Y_q = V(K_{h\times n}^{(k)})$. By the duality, we know that

is a transversal of $(K_{h\times n}^{(k)})^*$. Let $\mathscr{F} = \{F_1, F_2, \cdots, F_{|\mathscr{F}|}\}$. Then, $\mathscr{F}(h) = \{F_1(h), F_2(h), \cdots, F_{|\mathscr{F}|}(h)\}$ is a family of transversals of $(K_{h\times n}^{(k)})^*$. Obviously, $r_{\mathscr{F}(h)} = r_{\mathscr{F}}$. Hence, we have

for $h\geq 2$ and $k\geq 2$.

4.   Concluding remarks

By Theorems 3.2 and 3.3, we have completely determined the fractional total domatic number on the incident graph of $K_{h\times n}^{(k)}$ for all positive integers $n, k, h$.

Theorem 4.1. Let $n, \ k, \ h$ be positive integers, $n\geq k$. Then,

When $k = 2$, we simply denote $K^{(k)}_{h \times n}$ by $K_{h\times n}$. Recall that the incidence graph of a graph $G$ is exactly the subdivision graph $S(G)$. Then, we have the following result, which extends Theorem 1.1.

Theorem 4.2. For integers $n\geq 3$, $h\geq 1,$

As discussed in Lemma 3.1, for an arbitrary hypergraph $H$, the open neighborhood hypergraph $ON(I(H))$ of its incident graph $I(H)$ consists of two components: $H$ and its dual hypergraph $H^*$. By Lemmas 1.3 and 1.5, there is $FTD(I(H)) = DT_f(ON(I(H))) = \min \{DT_f(H), DT_f(H^*)\}\leq DT_f(H).$ In this paper, we have proved that $FTD(I(H)) = DT_f(H)$ when $H$ is an $h$-balanced $n$-partite complete $k$-uniform hypergraph for any positive integers $h, n, k$ ($n\geq k$). It is interesting to determine the class of hypergraphs $H$ with $FTD(I(H)) = DT_f(H)$.

Acknowledgments

This research is supported by the National Natural Science Foundation of China (No.12071265) and the Shandong Provincial Natural Science Foundation (No. ZR2019MA032).

Conflict of interest

All authors declare no conflicts of interest in this paper.