It would be bad of Cree if pot would need to be 10KOhm, it would mean that internal resistor is just 1kOhm, 10mA current flowing from sense wire to ground when set to minimum dimming (wires connected directly or pot turned all the way down). I think they would like to keep it around 1mA top but you never know.
If OP is up to it he can find exact value of internal resistor.
Get 100kOhm pot and multimeter.
Measure voltage of sense wire when disconnected (multimeter between sense wire and ground wire of dimmer input). It should be ~11V
Torn pot somewhere to the middle, measure its resistance between lets call it left pin and middle one. Adjust pot to about 25kOhm (you can use resistor with similar value, I would use 20-50kOhm).
Connect the same pins to dimmer input.
Measure voltage across them.
Now you have Rp (resistance of pot), Vi (input voltage to divider measured in step 1 when sense wire is disconnected and current is not flowing thru internal resistor), and Vp (voltage across pot).
From this internal resistor value Ri is Rp/(Vi/Vp-1)
Once you know Ri best pot would be one with 10 times bigger resistance or little more (assuming open sense pin has 11V, if it is closer to 10V then dimming pot needs to be x times bigger that difference between open pin voltage is bigger than 10V, e.g., if open pin voltage is 10.5V then pot needs to be 20 times bigger resistance then internal resistor), logarithmic preferably (because one part of divider is constant and linear will not change Vp quickly enough to see difference).
