**Conjecture**If are simple finite graphs, then .

Here is the tensor product (also called the direct or categorical product) of and .

This beautiful and seemingly innocent conjecture asserts a deep and important property of graph coloring. It is undoubtedly one of the most significant unsolved problems in graph coloring and graph homomorphisms.

We write if there is a homomorphism from to . The graph has a two natural projection maps (projecting onto either the first or second coordinate), and these maps are homomorphisms to and to . So, in short, and . A graph is -colorable if and only if it has a homomorphism to . Combining this with the transitivity of we find that (indeed, if , then and , so - equivalently, is -colorable). So, the hard direction of Hedetniemi's Conjecture is to prove that .

Let's define to be the proposition that whenever and . Then the above conjecture is equivalent to the statement that holds for every positive integer . Now holds trivially and follows from the observation that the product of two graphs each of which contains an edge is a graph which contains an edge. The next case is quite easy too, if and , then both and contain an odd cycle. Since the product of two odd cycles contains an odd cycle, this shows . The next case up, was proved by El-Zahar and Sauer by way of a beautiful argument. It is open for all higher values.

A key tool in the proof of El-Zahar and Sauer is the use of exponential graphs. For any pair of graphs the exponential graph is a graph whose vertex set consists of all mappings . Two vertices are adjacent if is an edge of whenever is an edge of . It is easy to see the relevance of to this problem. If we have an -coloring of , then for every vertex , there is a mapping given by . This associates each with a vertex in . Now it is easy to verify that whenever are adjacent vertices in , the maps and are adjacent in . Rather more surprisingly, Hedetniemi's conjecture may be reformulated as follows:

**Conjecture (version 2 of Hedetniemi)**If , then is -colorable.

The following conjecture asserts that Hedetniemi's conjecture still holds with circular chromatic number instead of the usual chromatic number. Here is the circular chromatic number of . Since this is a generalization of the original conjecture.

**Conjecture (Zhu)**If and are finite simple graphs then .

A graph has circular chromatic number for positive integers if and only if has a homomorphism to the graph . This is a graph whose vertex set consists of vertices cyclically ordered, with two vertices adjacent if they are distance apart in the cyclic ordering. So again, we may state this conjecture in terms of homomorphisms to graphs of the form . More generally, let us call a graph *multiplicative* if implies either or . Now Hedetniemi's conjecture asserts that every is multiplicative and Zhu's conjecture asserts that every is multiplicative. With this terminology, El-Zahar and Sauer proved that is multiplicative. A clever generalization of their argument due to Haggkvist, Hell, Miller and Neumann Lara showed that every odd cycle is multiplicative. Recently, Tardif bootstrapped this theorem with the help of a couple of interesting operators on the category of graphs to prove the is multiplicative whenever . Ignoring trivial cases and equivalences, these are essentially the only graphs known to be multiplicative.

It might be tempting to hope that all graphs are multiplicative, but this is false. To construct a non-multiplicative graph, take two graphs with the property that and (for instance and the Grotzsch Graph). Now is not multiplicative since and , but . It seems that there is no general conjecture as to what graphs are multiplicative. Some other Cayley graphs look like reasonable candidates to me (M. DeVos), but I haven't any evidence one way or the other.

Poljak and Rodl defined the function . So, Hedetniemi's conjecture is equivalent to . Using an interesing inequality relating the chromatic number of a digraph to the chromatic number of a type of line graph of , they were able to prove the following quite surprising result: Either is bounded by or .

There are a number of interesting partial results not mentioned here, and the reader is encouraged to see the survey article by Zhu.