Is there any connection between the computer science phrase "network topology" and the mathematical notion of a topological space (or, is there any other way to connect "network topologies" with formal definitions from the field of topology)?

Example: I suppose that if one said "there are two nodes A and B, and they are each connected to a third node C, and C is connected to a fourth node D, and D is node E and D is also connected to node F, and the names of A and B are "A" and "B", and the name of C is 'LAN router' and the name of D is 'ISP AP' and the name of E is 'Website 1' and the name of F is 'Website 2'", then the mathematical name for that sort of structure would be a "graph with labeled nodes". In addition, if one assigned a directionality to each link, eg "F > D, E > D, D > C, C > A, C > B", then this would make it a "directed graph with labeled nodes". And if some nodes had multiple ports, and each node gave locally unique names to each port, that would induce a unique naming on edges (the name of an edge is the 4-tuple (source node's name, source node's port, destination node's name, destination node's port)), and you could call it a "directed graph with labeled nodes and edges". All of these appear to me to fit within the computer science term "network topology", and all of the mathematical terms given above which appear to be from "graph theory". I am just unclear if these have anything to do with the mathematical area of "topology".

Also, does the phrase "network topology" have anything to do with "topological graph theory" or with abstract simplicial complexes?

Possibly related:

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    Any graph with, say, a finite number of edges at each vertex, can be made into a metric space by taking each edge to be length $1$ and the distance between any two vertices the length of the shortest path. Still, those uses of the word topology should not be thought of as mathematically valid. Nor would the computer people care either way. – Will Jagy Jan 25 '15 at 0:20
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    It depends who is using the term, but I have only seen used as more of less a synonym for "shape", i.e., with no real connection to the mathematical meaning. – Chris Godsil Jan 25 '15 at 1:15

Echoing the comment made by @Chris Godsill, in this usage the "topology" of a network is indeed its "shape". However, I disagree with his claim that there is no clear and relevant connection to topology in its formal mathematical sense. This is because whenever mathematicians talk about the "topology" of a space, we really are talking about the kind of shape it has!

Of course, the whole point of the mathematical field of topology is to make our various intuitive notions of "shape" precise. As @MJD has pointed out, probably the most pertinent topological property of a network is its connectedness (or to be more specific in this case, path-connectedness), which tells us which nodes are connected to which other ones. A network in which any node can communicate with any other e.g. in a star or ring topology (connected), is fundamentally different from one in which one lone node is unplugged from the rest, or one with two separate components $A, B$, where nodes are restricted to communicate within their own group (disconnected).

We can take this even further. Another important property of a network is how robust it is: what will happen if one of the cables connecting some nodes goes down? In a ring topology, communication between any pair of nodes is still possible even upon the disconnection of one of the links, since we may send the data "the other way round" the network. Contrast this with the star topology, where all data must pass through the central hub, so losing any of the spokes means that node is no longer in communication with the rest of the network. We formalize this in graph theory (which is basically almost a branch of topology anyway :P) as the edge-connectivity of the network -- the minimum number of edges we have to remove before the space becomes disconnected. The star topology is $1$-edge-connected, we only have to remove one edge to disconnect it, while the ring topology is $2$-edge-connected. The fully-connected mesh shown in john mangual's answer is $(k-1)$-edge-connected, where $k$ is the number of nodes.

You mention simplicial complexes: these provide the link between graph theory and topology you were seeking. Any network is, as you already know, a graph. But a graph is just a simplicial 1-complex, where the nodes are the 0-simplices, and the edges are 1-simplices. So any of the methods we can use on simplicial complexes can be used on graphs.

Here's one example, it's a bit long and isn't entirely necessary to the overall answer, so feel free to skip this paragraph. On a related (but less pragmatic) note to the previous paragraph, observe that the ring topology is different from the star topology in the sense that it's a "circle", i.e. it has some sort of a "hole" that the star does not. So holes play an important role in determining the shape of a space too. This intuitive idea is formalized in algebraic topology using the fundamental group $\pi_1$, which fixes a basepoint $x$ in a simplicial complex $\mathscr{K}$ and looks at all loops at $x$, up to homotopy (intuitively, two loops are homotopic if we can deform one into the other, all the while remaining inside the space $\mathscr{K}$). Note that a loop is allowed to cross over itself, all that is required is that it starts and ends on $x$. Say we fix the basepoint $x$ of the star graph at its central node. What are the loops? Well, if we start at $x$ and go along any of the spokes, we can only come back to $x$ by coming back along the same spoke. I hope it's then intuitively obvious that such a loop can be deformed by "pulling it in" towards $x$, and so all loops in the star are homotopic to the trivial loop, that starts at $x$, stays at $x$ and ends at $x$. In a sense, there's only one "loop" in the star, and it's not really a loop anyway. In contrast the ring network, in addition to having these stupid loops, also has the "proper" loop that goes around the ring once clockwise. But it also has the loop that goes around twice, three times, ..., $n$ times clockwise where $n \in \mathbb{Z}$ ($n < 0$ just means we go around the loop anti-clockwise). It's possible to show that none of these loops deforms into any other while remaining in the ring, so there are $\mathbb{Z}$ many loops in the fundamental group of the ring. I haven't explained how we define the group operation, but once we do this we see that $\pi_1$ for the star is trivial i.e. the single-element group $\{0\}$, while $\pi_1 = \mathbb{Z}$ for the ring. In general, the "number of copies of $\mathbb{Z}$" in the fundamental group tells us how many (1-dimensional) "holes" the simplicical complex has, so as expected, our result tells us that the star has no holes, while the ring has one.

tl;dr, considering graphs as simplicial 1-complexes allows us to use algebraic topology to probe properties of the shape of the network. So yes, network topologies, interpreted the right way, are topological spaces.

I expect this is how much of topological graph theory goes too, I don't know much about it yet. One other important topological property I can think of is the graph distance and the diameter of the network, one would likely consider this in data routing efficiency problems. Example, the star is good in the sense that data need only ever travel the length of two links to get from any one node to another, while the ring may have to shuttle it over many links if the nodes are "far apart" in the network. Mathematically, the diameter of the star is 2, while that of the ring is $k/2$ where $k$ is the number of nodes.

  • Also late but: my understanding is that "topological graph theory" is sort of a euphemism for a generalization of "planar graph theory" when you embed into other topological spaces (like face-colorings in surfaces, or knottedness in 3-manifolds). – Eric Stucky Mar 8 '16 at 17:31

The commonality is this: In topology we don't care about where things are or how big they are; we care about what is connected to what. A square and a circle are topologically equivalent, because they are self-connected in the same way. You can stretch the square and smooth out the corners and make it a circle and the topology doesn't care. But if you disconnect two of the sides, even a tiny bit, suddenly it's a different thing.

In computer networks, we similarly don't care (much) about where things are or how long the connections are. But we care a lot about which computers are connected to which. You can pick up a computer on the network and carry it to the other side of the room and it will go on communicating as if nothing had happened, because from the point of view of the network, nothing has happened. But if you reach in back and pull the network cable out by a thousandth of an inch, suddenly it's a different network.

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    You had a lot of other questions, but I didn't want to clutter up my answer addressing them. Contrary to the comments, I think the computer science invocations of topology are mostly accurate. When we talk of network topology, it is in contrast to network geometry, and we are signaling that, like topologists, we don't care (for the moment) were anything actually is. – MJD May 12 '15 at 2:45

Network topology is pretty much what we would call graph theory and notions from the different areas line up.

Now graphs do inherit a topology and even a metric in the mathematical sense, if you want to go there.

star network = star graph

Fully Connected Mesh network = Complete Graph

Ring network = Cycle graph

  • Sorry, can't resist the urge to "go there". Starting with the notion of neighborhoods as a set containing a node and all directly connected vertices ($\{i\}\cup\{j:(i,j)\in E\}$), and then flushing this basis out to a topology by including intersections and unions, I get the following: – cantorhead Jan 15 '17 at 16:27
  • Cycle graph = discrete topology; Complete Graph = trivial topology; star graph = an interesting non-Hausdorff topology (I think $T_0$ or $T_1$ but not $T_2$). Is this correct? – cantorhead Jan 15 '17 at 16:27
  • @cantorhead why is it not hausdorff? i always have trouble with this concept – cactus314 Jan 15 '17 at 16:47
  • Hi. I believe that a given peripheral node cannot be separated from the center node in the Hausdorff sense, meaning that there are no disjoint open sets $U_1$ and $U_2$ containing the center and peripheral nodes respectively... – cantorhead Jan 15 '17 at 17:09
  • Let 1 be the center node and 2 through 6 be the peripheral nodes. Then $\{1,2\}$ is an open set since there is an edge. But, every open set containing 2 will also contain 1. This is because among the basis open sets, $\{1,j\}$, $j=2,\ldots,6$ they all contain 1 so that no matter what intersection you take the only way to end up without 1 is the empty set. – cantorhead Jan 15 '17 at 17:09

Probably a bit late, but I asked my physics professor about the relationship between Mathematical Topology and network topology, or what he mentioned as topology, the different basic circuit diagrams. He said that electricians, network engineers, and mathematicians look at it differently, but it is the same kind of topology.

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