In steady state climb or descent the vertical component of lift is always 1g - you only need more or less than that to initiate the climb or descent.
I agree that in a steady-state climb or descent, the sum of the vertical forces is zero. In a straight, non-turning descent, the downwards force of weight is balanced by the upwards components of the lift force and the drag force. (So I would actually say that the vertical component of lift is less than 1g; the balance being provided by the vertical component of drag).
But we are talking about a turning descent, during which there must be an unbalanced force causing the aircraft to accelerate towards the centre of the turn. This force is provided by a horizontal component of the lift. So while the vertical component of lift (together with the vertical component of drag) is balancing the weight, the extra lift produced by 'pulling g' is used to accelerate the aircraft around the turn. Therefore, the total lift vector is more than the weight (or more than 1g).
Or have I got this all wrong somewhere?