Basic ternary circuit to light three LED’s?

Nice simple circuit with two relays and two diodes, though I'm sure this is what you currently have. I don't know that a design with more parts will count as simpler.

I have made the assumption that the control signal can source enough current to light the LEDs. I also decided that the LEDs should be off when the input is floating. I then created the circuit shown below.

With a floating input Q1 is off and the current in R4 flows through D1 and D2 and makes the voltage at point A about 4V. When the input is connected to ground, Q1 turns on and LED3 lights and the the voltage at point A drops to about 3V. As the input goes negative LED1 starts to conduct ; at about -10V all the current from R4 flows through LED1. LED3 will be off and the voltage at point A will be around zero volts. When the input goes positive Q1 and LED3 turn off and LED2 turns on; the voltage at point A will stay around 4V. Since the circuit draws current from V+ no matter what, putting a fourth LED in series with R4 would prevent users leaving power on. If you don't have a 3.3V zener D1 and D2 can be LEDs; they can be placed out of sight or used as power on indicators. The resistor values are the ones I used; you may need to scale them to suit your components.

Here is a 4-transistor + 3 diode solution.

V2 +12 D1 on 10mA only

V2 -12 D2 on 10mA only

V2 0V D3 on 10mA only

V1 0V all off.

^{simulate this circuit – Schematic created using CircuitLab}

To minimize the current drawn would require another transistor to invert the drive to D3 rather than shunting it.

There might be a simpler approach, this does not quite feel elegant enough.

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Edit:

Okay, using Jasen's idea of common-base input, the below gets rid of the diodes and still has 4 transistor (5 if you want to avoid the shunt)

^{simulate this circuit}

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A couple of opto-isolators can be used to detect polarity and give logic level outputs.

You'll need a little additional logic for the red lamp.

something like this could do it.

^{simulate this circuit – Schematic created using CircuitLab}