How can I "open" a file from WSL with the default application?

Since the Windows Linux interop started working you can now call:

cmd.exe /C start <file>

As Martijn noted this is the correct way to execute/open a Windows application/file.

cmd.exe /C start <file>

I found it very useful to work this into a bash script that I keep in a folder that is in my system path. I name it start and do chmod 0744 to the file to make it executable. The $* means it will pass all of the command line arguments you provided to the script to cmd.exe.

#!/bin/bash
cmd.exe /c start "Launching from BASH" "$*"

With this command in my system path I can commands like this in Linux that open in Windows:

1. start FileXYZ.pdf // Opens the PDF in the default assigned PDF viewer in Windows
2. start explorer . // Opens current WSL folder in the Windows Explorer
3. start MyApp.exe // Launches the Windows application

to expand on Martijn's answer, you can put

alias start='cmd.exe /C start'

in your .bashrc to get expected windows behavior, eg start . opens explorer in current dir.

This worked much better for me:

explorer.exe `wslpath -aw <path>`