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I think this is more stackexchange-worthy, but here goes.

I'm a little afraid you're mixing up general weight diagrams with the root system, which is the weight diagram of the adjoint representation. The root system of $G\times H$ lives in a space that's just the Cartesian product of the two individual spaces, and is the disjoint union: $$\Delta_{G\times H} = (\Delta_G \times 0) \cup (0 \times \Delta_H).$$

If one's talking about weight diagrams of irreducible representations of $G\times H$ (which the adjoint representation is not), it helps to know that they're all of the form $V\otimes W$. So then, one takes the convolution of the two weight multiplicity diagrams, again inside that Cartesian product:

$$mult(V\otimes W){\lambda,\mu} W)_{\lambda,\mu} = mult(V)\lambda mult(V)_\lambda \ mult(W)_\mu.$$

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I think this is more stackexchange-worthy, but here goes.

I'm a little afraid you're mixing up general weight diagrams with the root system, which is the weight diagram of the adjoint representation. The root system of $G\times H$ lives in a space that's just the Cartesian product of the two individual spaces, and is the disjoint union: $$\Delta_{G\times H} = (\Delta_G \times 0) \cup (0 \times \Delta_H).$$

If one's talking about weight diagrams of irreducible representations of $G\times H$ (which the adjoint representation is not), it helps to know that they're all of the form $V\otimes W$. So then, one takes the convolution of the two weight multiplicity diagrams, again inside that Cartesian product:

$$mult(V\otimes W){\lambda,\mu} = mult(V)\lambda \ mult(W)_\mu.$$