If the direct product of two semilattices exists, what does its Hasse diagram look like in terms of its constituent semilattice Hasse diagrams?

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This is likely to be a quick question.

Definition: A semilattice $(L,\lor)$ is a commutative, idempotent semigroup.

The Hasse diagram $H$ of $L$ is with respect to the order $x\le y$ iff $x\lor y=y$.

The direct product of two semilattices would be, I suppose, in the sense of universal algebra.

The Question:

Suppose $L_1, L_2$ are semilattices with Hasse diagrams $H_1,H_2$, respectively. What does the Hasse diagram $H$ of $L_1\times L_2$ look like (and how does it relate to $H_1$ and $H_2$)?

Thoughts:

Is it as simple as juxtaposition, with $H_1$ on the left of $H_2$, or are they connected, like how two Necker cubes form a tesseract in 2D (so to speak)? (I won't insult your intelligence by drawing them or describing these thoughts in more detail.)

What is causing the confusion?

I don't have a clear picture (pun intended) of what the direct product is in this context: sure, the underlying set is $L_1\times L_2$, but is the operation

$$(a,b)\lor(x,y)=(a\lor_1 x,b\lor_2 y)?$$

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Yes, the direct product is equipped with operation $$(a, b)\vee (x, y)= (a\vee x, b\vee y).$$ This is the product of those semi-lattices as universal algebras, as well as the categorical product.

The order on $L_1\times L_2$ is \begin{align*}(a, b)\leq (x, y) \iff & (x, y)\vee (a, b) = (x, y) \\ \iff & x = x\vee a \text{ and } y = y\vee b \\ \iff & a\leq x\text{ and }b\leq y.\end{align*}

$L_1\times L_2$ equipped with this partial order is called the product of posets $L_1$ and $L_2$.

Hasse diagram of such poset is a bit hard to imagine since going just from the definition of the product order, its four dimensional. Restricting ourselves to $L_2$ a linear order, for example $L_2 = \{1, \cdots, n\}$ with standard ordering, we can graph the order of $L_1\times L_2$ in three dimensions by a natural grading coming from $L_2$, drawing $H_1$ on the $xy$-plane, and putting a copy of it at heights $z = 1, ..., n$, with arrows from $(x, i)$ to $(x, i+1)$ for $x\in L_1$ and $1\leq i \leq n-1$. Of course, to obtain a proper two dimensional Hasse diagram of $L_1\times L_2$ we need to rearrange the directed graph onto the $xy$-plane again.

For example, if both $L_1$ and $L_2$ are linear orders, we have a ready-to-go visualization for the Hasse diagram of $L_1\times L_2$, without the need to reduce this graph from three to two dimensions.