Can AES GCM mode detect an incorrect key and refuse to decrypt? I understand that AES in GCM mode produces an authentication tag that is used to ensure that a tampered with cipher text is not decrypted. However, if I pass the wrong key will AES in GCM mode detect that I am using the wrong key and give me an error message or will it give me back garbled data because I used the wrong decryption key?
Does the AES-GCM spec say that an error will be raised if the wrong key is used?
UPDATE The error does not have to indicate that the key was wrong, just that decryption should fail so incorrect clear text is not returned to the caller.
An AES-GCM-encrypted message is obtained per an encryption key $K_0$. As for any correct implementation of AES-GCM, we assume that legitimate encryption of multiple ciphertexts with $K_0$ uses $\text{IV}$ such that the counters (obtained by increment of the $\text{IV}$) do not overlap; and that the implementation of decryption gives an error message when the authentication check fails.

will AES in GCM mode detect that I am using the wrong key and give me an error message (..) ?

No, with strictly only what's stated. It is possible to exhibit a ciphertext (including $\text{IV}$) and two distinct keys $K_0$ and $K_1$ that both decipher successfully (and a plaintext that, encrypted with key $K_0$ and $\text{IV}$, encrypts to the ciphertext including $\text{IV}$ and authenticator tag, thus matching the wording in the above re-statement of the question).
But yes, if the ciphertext was prepared legitimately (which was my assumption in a first version of this answer). An attack against a given ciphertext, aiming at exhibiting $K_1$, has cost $p\,2^k$ AES encryption for probability $p$ of success and $k$-bit authentication tag, even if multiple ciphertexts are available. Prepared legitimately holds at least if one (or both) of the following holds:

The key $K_0$ is chosen uniformly at random independently of $\text{IV}$ and the plaintext (even if the plaintext, $\text{IV}$, and $K_1$, are chosen maliciously). That includes any case where $K_0$ is kept secret from adversaries at least until production of the plaintext, which is a standard assumption.
The $\text{IV}$ is chosen uniformly at random independently of $K_0$ and the plaintext (as is the case in some but not all legitimate uses of AES-GCM encryption); that's even if the plaintext, $K_0$, and $K_1$, are chosen maliciously.

Argument: from the structure of GCM it can be shown, under a model of AES as an ideal cipher, that the AES-GCM authenticator tag is a PRF of the key and $\text{IV}$, for any fixed message (plaintext, or ciphertext excluding $\text{IV}$ and authenticator tag).

Conclusion: AES-GCM encryption with secret key $K_0$, or with random unpredictable $\text{IV}$, makes it infeasible that later (including if $K_0$ leaks) anyone could find (much less encounter by accident) another key $K_1$ that allows decryption of the unmodified ciphertext (including $\text{IV}$ and authenticator), yielding almost certainly corrupted deciphered plaintext. However, that can be defeated by the party choosing $K_0$/doing the encryption/choosing the message; and the essential unmodified ciphertext hypothesis is artificial, for it goes against the rationale of using authenticated encryption.