How to know if a Feynman diagram is planar? A planar diagram is defined as being one of the leading diagrams for $N \to \infty$ (large $N$ expansion), and, as I understand it, it should have the lowest genus when compared to a non-planar diagram. It is of course very useful to be able to distinguish planar from non-planar just by looking at the diagram without having to compute color factors every time.
To illustrate my misunderstanding, let us consider the two following diagrams:

Here the circles top and bottom correspond to trace, i.e. these diagrams represent a two-point function with composite operators of the form:
$$\mathcal{O}_k (x) \propto \text{Tr}\ T^{a_1} T^{a_2} T^{a_3} T^{a_4} \phi^{a_1} \phi^{a_2} \phi^{a_3} \phi^{a_4}, \tag{1}$$
with the $a$'s corresponding to color indices and the $\phi$'s are scalar fields.
I would expect diagram $(2)$ to be non-planar, however when I do the computation for each of them I find that both diagrams have the same planarity, since:
$$\begin{align} \text{Tr}\ T^a T^b T^c T^d\ \text{Tr}\ T^a T^b T^c T^d & = \frac{1}{2} \text{Tr}\ T^a T^b T^c T^a T^b T^c \\ &= \frac{1}{4} \text{Tr}\ T^a T^b\ \text{Tr}\ T^a T^b \\ & = \frac{1}{8} \text{Tr}\ T^a T^a \\ &= \frac{N^2}{16}, \end{align} \tag{2.a}$$
and
$$\begin{align} \text{Tr}\ T^a T^b T^c T^d\ \text{Tr}\ T^b T^a T^c T^d & = \frac{1}{2} \text{Tr}\ T^a T^b T^c T^b T^a T^c \\ &= \frac{1}{4} \text{Tr}\ T^a T^b\ \text{Tr}\ T^b T^a \\ &= \frac{1}{4} \text{Tr}\ T^a T^b\ \text{Tr}\ T^a T^b \\ &= \frac{N^2}{16}, \end{align} \tag{2.b}$$
where I made heavy use of the following identities:
$$\text{Tr}\ T^a A\ \text{Tr}\ T^a B = \frac{1}{2} \text{Tr}\ A B, \tag{3.a}$$
$$\text{Tr}\ T^a A T^a B = \frac{1}{2} \text{Tr}\ A\ \text{Tr}\ B. \tag{3.b}$$
Is that right? If yes, how can I distinguish planarity diagrammatically then?
 A: There is celebrated formula for Euler characteristic $$\boxed{\chi = V - E + F}$$  where $V$ is number of vertices, $E$ - number of edges, $F$ - number of faces involved in the graph. The Euler  characteristic is related to the genus $g$ of surface by $\chi = 2 - 2g$. The planar graph corresponds to sphere with $g= 0$. 
For the graph (1) you have 2 vertices, 4 edges, 4 faces (exterior face has be taken also into account) - therefore, it is a planar graph. For the graph (2) there seem to be 2 faces, so this graph can be put only on torus, or higher genus surface. 
A: Concerning OP's diagram (1) & (2) and OP's calculations, note that the labelling of the second vertex is reversed, i.e. the color factor becomes 
$$  {\rm Tr}(T^a T^b T^c T^d) {\rm Tr}(T^d T^c T^b T^a)~\stackrel{(3.a')+(3.b')}{=}~({\rm Tr}\mathbb{1})^4+\text{subleading terms}, \tag{2.a'}$$
and 
$$  {\rm Tr}(T^a T^b T^c T^d) {\rm Tr}(T^d T^c T^{\color{red}{a}} T^{\color{red}{b}})~\stackrel{(3.a')+(3.b')}{=}~({\rm Tr}\mathbb{1})^3 +\text{subleading terms},\tag{2.b'}$$
respectively. We see that diagram (2) has less index contractions [i.e. factors of ${\rm Tr}\mathbb{1}=N$], which is a hallmark of a non-planar diagram.
Here we have repeatedly used the formulas
$${\rm Tr}( T^a A) {\rm Tr}( T^a B)~=~ {\rm Tr}(A B)+ \text{subleading terms}, \tag{3.a'}$$
and
$${\rm Tr}(  T^a A T^a B) ~=~   {\rm Tr}(A){\rm Tr}(B)+\text{subleading terms}. \tag{3.b'}$$
References:


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*D. Tong, Gauge theory lecture notes; chapter 6.

