I am having problem understanding what negative of a combinatorial game $G$ exactly means in combinatorial game theory. Does it mean that if I have normal game, if I create inverse, i.e., $G = \{G^R \mid G^L\}$ I get a Misere game?

$\begingroup$ Here's a concrete example to keep in mind: In (RedBlue) Hackenbush, taking the negative of a game simply amounts to color swapping. There are several ways to see that this is generally different from "miserification." The simplest is probably just computing a specific example, but more generally note that no Hackenbush game is a firstplayer win. However, plenty of misere Hackenbush games (e.g. the misere version of one red stalk + one blue stalk) are firstplayer wins. $\endgroup$– Noah SchweberSep 19 at 2:58
2 Answers
No, the negative a game is simply the game in which the player's roles are swapped, hereditarily. You can see this in the definition you provided $$G=\{ G^R\mid G^L\}$$ since the left options in $G$ are the (negations of) the right options in $G$ and vice versa.
If we take the sum of a game $G$ and its negative $G$, we get a game in which any move by any player can be copied in the other part of the game, which implies $G+(G)=0$.
The negative of a game is not the same as the misère game, since in $G$ it is not that you have the same options as before and are trying to lose, but rather that you have your opponent's options and are trying to win as though playing as them.
For example, the negative of a game of chess is still very much a game of chess—you are simply playing now as Black instead of White. But misère chess looks totally different in the manner of play, since players are exposing their queens and so forth in attempt to lose.

$\begingroup$ Misere chess is certainly very different. But the players are probably not "exposing their queens." To win, you still have to control the game. A strong misere player playing a weak misere player will probably gobble up material, then almost paralyze the weak player so that the weak player has only one or two legal moves on each turn. Only in that way can the weak player be forced to checkmate the strong player. Selfmate puzzles may give some sense of what the end of a misere chess game looks like. $\endgroup$ Sep 19 at 2:00

$\begingroup$ @TimothyChow Some forms of misère chess play with a forced capture rule (if you can capture you must), and in these forms, it can be advantageous to expose and thereby get rid of pieces that otherwise will be in effect controlled by your opponent. $\endgroup$ Sep 19 at 2:22

$\begingroup$ But I agree that without that convention play would be different. $\endgroup$ Sep 19 at 2:42

1$\begingroup$ @JoelDavidHamkins I think that version of chess, while much more natural to play, is not strictly "misere chess" in the gametheoretic sense. $\endgroup$ Sep 19 at 2:55

$\begingroup$ Yes, agreed. I was thinking of it mainly because those are the rules by which misere chess is often played in the club rooms at scholastic chess tournaments I have often attended when my kids were younger. $\endgroup$ Sep 19 at 3:07
Others have already pointed out to you that the answer is negative. Let me add the following take: if you imagine a game as a bicolored tree (where the root is the starting position, and the edges are colored red or blue according as they are traversable by the right or left player, and where we play following the normal “who cannot play has lost” rule):
To form the negative of a game, you simply exchange the colors of the edges.
To form the misère of a game, you take each node which has no red (resp. blue) child (i.e., no red, resp. blue, edge leading away from the root) and you add a new red (resp. blue) edge from that node leading to a new terminal node¹, allowing the player who would have lost in the normal version of the game to declare victory in the misère form. That new node ensures that the opposite player has lost, thus “converting” the misère convention to the normal convention.
A concrete example where the two constructions differ is a typical impartial game (“impartial” meaning both players have identical options in any position): the negative of an impartial game will be the game itself, whereas the misère will typically differ. In the game of nim, for example, while the misère's version strategy is the same for most of the game, it differs at the end: clearly if there is only one row of sticks left, in the normal version the winning move is to take them whereas in the misère version you leave one, forcing your opponent to take it.
Another thing that must be pointed out is that the negative of a game is compatible with Conway equality (Conway equality of games $G\doteq H$ means that $G + (H)$ is a secondplayer win): that is, if $G\doteq G'$ then $(G) \doteq (G')$, so “negative” is a welldefined construction on the game values. On the other hand, “misère” depends on the specific form of a game, it is not defined on the game values: that $G\doteq G'$ does not imply $\operatorname{mis\grave ere}(G) \doteq \operatorname{mis\grave ere}(G')$, as evidenced by the fact that the nim positions $1+1$ and $2+2$ are both secondplayer wins in normal play (so they have nim value $0$: they are Conwayequal) whereas in misère play the first is a firstplayer win and the second is a secondplayer win.
 Maybe that node should have a red (resp. blue) edge leading back to itself? This wouldn't change the game itself, but it changes the way it gets added to other games. The misère construction isn't entirely wellspecified (as far as I know).

$\begingroup$ No, leading to a terminal node. Otherwise the misère version of the plain zero game would be dud which is a draw. $\endgroup$ Sep 24 at 14:08

$\begingroup$ @NoLongerBreathedIn I mean edges leading back to itself and accessible only from the player for whom the terminal node is supposed to guarantee a win (in Winning Way's terminology this would be on or off, and the misère version of the zero game would be on  off). My point is, there is no clear convention on how to add misère games. $\endgroup$– GroTsenSep 25 at 8:18

$\begingroup$ I feel like the misère form of a nonloopy game should still be nonloopy, but it might not make a huge difference. $\endgroup$ Sep 28 at 4:37